ANSWER:
0.011 J
STEP-BY-STEP EXPLANATION:
Given:
q = 150 μC
v = 75 V
r = 0.2
We calculate the work as the product between the charge and the potential difference, just like this:
[tex]\begin{gathered} W=q\cdot v \\ W=150\cdot10^{-6}\cdot75 \\ W=0.01125\text{ J} \\ W\cong0.011\text{ J} \end{gathered}[/tex]The work required is 0.011 J
if an astronaut weighs 981 N on Earth and only 160 N on the moon, then what is the mass on the moon
If an astronaut weighs 981 N on Earth and only 160 N on the Moon, then his mass on the Moon will be 98.1 kg.
Let's calculate the mass of Earth as per the Earth's acceleration due to gravity.
Now, considering the acceleration due to gravity as 10m/s².
Mass = Weight/Acceleration due to gravity
Mass = 981/10
Mass = 98.1 kg
Now, as per the established fact, the mass is independent of acceleration due to the gravity of the planet i.e. The mass of the person on earth and on the moon is same.
Hence, his mass on the Moon will be 98.1 kg.
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When light goes from a slower medium to a faster medium; what way does the light bend relative to the normal?
Answer: A.
It moves away from the normal.
Explanation: ed mentum or plato
When doing a chin-up, a student lifts her body with a force of 400N and a distance of 0.25 meters in 2 seconds. What is the power delivered by the students biceps?
Given:
The student lifts her body with a force F = 400 N
The distance is d = 0.25 N
The time is t = 2 s
To find: The power delivered by student's biceps.
Explanation:
The formula to calculate power is
[tex]P=\frac{F\times d}{t}[/tex]Substituting the values, the power will be
[tex]\begin{gathered} P=\frac{400\times0.25}{2} \\ =\text{ 50 W} \end{gathered}[/tex]Final Answer: The power delivered by student's biceps is 50 W.
Which of the following is not a unit of a force?
Answer: B
Explanation:
One of the common units of force is Newtons, N
Recall,
Force = ma
where
m is the mass of the body
a is the acceleration of the body
If acceleration, a = ms^-2 and mass = kg, then
Force = kgms^-2
Also, joule is the unit of work
recall,
work = force x distance = J
If we divide work by distance(meters), it becomes joule/meter
Thus,
jm^-1 is a unit of force
Thus, the option that is not a unit of force is
B. Js^-1
Two forces F1 = -6.00i + 7.90j and F2 = 6.80i + 5.30j are acting on an object with a mass of m = 4.10 kg. The forces are measured in newtons, i and j are the unit vectors. What is the magnitude of the object's acceleration?
The magnitude of object's acceleration is 3.26m/s².
The mass of the body is 4.10 lg.
The two forces that are acting on the object are F₁ = -6i + 7.9j newton and F₂ = 6.8i + 5.3j Newton.
We know that the force acting on an object is,
F = Ma
Where,
F is the force acting,
M is the mass of the object and,
a is the acceleration of the object.
As we can see, two forces are acting on the body,
We can simplify the forces in x direction and y direction,
The forces are F₁ = -6i + 7.9j N and F₂ = 6.8i + 5.3j N.
So, the total force in x-direction,
Fₓ = (-6+6.8)i
Fₓ = 0.8i
Fᵧ = (7.9+5.3)j
Fᵧ = 13.2j
So, the net force Fₙ on the object is Fₙ = (0.8i + 13.2j) N
Now, putting value of force and mass in the formula,
F = Ma
0.8i + 13.2j = 4.1a
a = 0.19i + 3.21j m/s².
The magnitude of acceleration is,
|a| = √[(0.19)²+(3.21)²]
|a| = 0.361 +10.3
|a| = 3.26m/s².
So, the magnitude of acceleration is 3.26m/s².
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A roller coaster car is traveling through a loop at 13 m/s. If the loop has a 23m radius, what centripetal force will the 53kg rider feel?
Given,
The velocity of the car, v=13 m/s
The radius of the loop, r=23 m
The mass of the rider, m=53 kg
The centripetal force is the force that keeps an object in its circular path.
The centripetal force is given by,
[tex]F=\frac{mv^2}{r}[/tex]On substituting the known values,
[tex]\begin{gathered} F=\frac{53\times13^2}{23} \\ =389.43\text{ N} \end{gathered}[/tex]Therefore, the centripetal force on the rider is 389.43 N
The x-component of a force on a 46-g golf ball by a 7-iron versus time is plotted in the following figure: a. Find the x-component of the impulse during the intervals i. [0, 50 ms], and ii. [50 ms, 100 ms] b. Find the change in the x-component of the momentum during the intervals iii. [0, 50 ms], and iv. [50 ms, 100 ms]
The x-component of the impulse during the intervals [0, 50 ms] is 750 Nms.
The x-component of the impulse during the intervals [50, 100 ms] is 1,500 Nms.
The change in the x-component of the momentum during the intervals [0, 50 ms] is 0.75 kgm/s.
The change in the x-component of the momentum during the intervals [50, 100 ms] is 1.5 kgm/s.
What is the impulse experienced by the ball?
The impulse experienced by the ball is calculated from the product of force and time of motion of the ball.
J = Ft
where;
F is the applied forcet is the time of motionThe x-component of the impulse during the intervals i. [0, 50 ms] is calculated as follows;
From the diagram, the impulse between (0, 50 ms) is the area of the triangle.
Jₓ = ¹/₂(b)(h)
where;
b is the base of the triangle = 50 ms h is the height of the triangle = 30 NJₓ = ¹/₂(50 ms)(30 N) = 750 N.ms
The impulse during [50 ms, 100 ms] is the area of the rectangle,
Jₓ = Lb
where;
L is the length = 100 ms - 50 ms = 50 msb is the breadth = 30 NJₓ = 50 ms x 30 N
Jₓ = 1,500 Nms
Impulse is the change in momentum of an object.
The change in the x-component of the momentum during the intervals [0, 50 ms] is calculated as follows;
ΔP = Jₓ = ¹/₂(50 ms)(30 N) = 750 N.ms = 0.75 Ns = 0.75 kgm/s
For interval of [50 ms, 100 ms];
ΔP = 1,500 Nms = 1.5 Ns = 1.5 kgm/s
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draw each of the following vectors, label an angle that specifies the vectors direction, then find its magnitude and direction.a. B= -4.0 I+ 4.0jb. r= (-2.0i-1.0j) cmc. v= (-10-100j) m/s d. a= (20i+10j) m/s^2the I's and j's have the hat and its the vector simble on the letters
A boy walks 4 km east and then turns around and walks 1.5 km west. If east is taken as the positive direction and the west as the negative direction on a number line then what is the distance the boy covers and his displacement?
ANSWER
[tex]undefined[/tex]EXPLANATION
First, let us make a sketch of the question:
From the diagram, the blue circle represents his starting position while the black circle represents his final position.
The total distance the boy covers is the sum of his two journeys, to the east and then to the west.
That is:
[tex]\begin{gathered} 4+1.5 \\ 5.5\operatorname{km} \end{gathered}[/tex]The boy's displacement is the distance between his starting position and his final position.
Since the west is taken as negative direction and the east taken as the positive direction, it means that 4km east means +4km and 1.5
Emily and Gemma did a Reaction time lab. Emily dropped the ruler while Gemma tried to catch it. She caught the ruler 5 times and her average catching distance is 0.12 m. What is Gemma's reaction time?
The average reaction time of Gemma is 0.1564 seconds.
As we know, Gemma is catching the scale and Emily is dropping the scale.
The whole experiment is taking place under gravity, so the acceleration is constant.
As we know, the scale is dropped, it means that the initial velocity of the scale is zero.
We can use the equation of motion,
The equation is,
S = Ut + 1/2at²
Where,
S is the displacement, which is 0.12 m in our case,
U is initial velocity which is 0m/s because the stone is dropped,
t is the time taken, this is equal tot he reaction time here,
a is the acceleration due to gravity whose value is 9.8m/s.
Now, putting all the values,
0.12 = 1/2(9.8)(t)²
t² = 0.24/9.8
t = 0.1564 seconds.
Gemma reacts in 0.1564 seconds to catch the scale.
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A cyclist travels a distance S from A to B consisting of three segments: s = S1 + $2 + $3, Time and speed over the segments are t1, t2, t3 and v1= 20(km/h), v2 = 15(km/h), V3= 10(km/h). Calculate the average speed over the distance s. Consider two cases: a) s1=s2=s3 b) t1=t2=t3
The average speed over the distance s consisting of three segments s1, s2, and s3 in first case a) s1 = s2= s3 is 13.85 km/h and in the second case b) t1 = t2 = t3 is 15 km/h.
In the first case where the distance for three intervals is the same a) s₁ = s₂ = s₃, the average speed for three segments is calculated by the formula:
Average speed = 3 ÷ ( 1/v₁ + 1/v₂ +1/v₃)
Average speed = 3 ÷ ( ¹/₂₀ + ¹/₁₅ + ¹/₁₀)
Average speed = 13.85 km/h
In the second case where the time taken for all three intervals is the same b) t₁ = t₂ = t₃, the average speed for three segments is calculated by the formula:
Average speed = (v₁ + v₂ +v₃) ÷ 3
Average speed = (20 + 15 + 10) ÷ 3
Average speed = 15 km/h
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A ball rolls off a table and falls 0.75m to the floor, landing with a speed of 4 m/s. (A) What is the acceleration of the ball just before it strikes the ground? (B) What was the initial speed of the ball? (C) What initial speed must the ball have if it is to land with a speed of 5 m/s?
Answer:
We had this question yesterday, let me check my book quickly
(A) The acceleration of the ball is 9.81 m/s² just before it strikes the ground.
(B) The initial speed of the ball is equal to 1.14 m/s.
(C) The initial speed must be 3.21 m/s if it is to land with a speed of 5 m/s.
What are the equations of motion?The equation of motion is the way to represent the relation between the time, acceleration, initial and final velocity, and distance covered by a moving object.
The three equations of motion are:
[tex]v= u+ at\\v^2 = u^2 +2aS\\S = ut +(1/2) at^2[/tex]
The acceleration of the ball just before it strikes the ground is equal to gravitational acceleration, g = 9.81 m/s².
Given, the final velocity of the ball, v = 4m/s
The height of the table, h = 0.75 m
The initial velocity of the ball, [tex]u = \sqrt{v^2-2gh}[/tex]
[tex]u = \sqrt{(4)^2-2\times 9.8\times 0.75}[/tex]
u = 1.14 m/s
When the final velocity of the ball, v = 5m/s
The initial velocity will be :[tex]u = \sqrt{(5)^2-2\times 9.8\times 0.75}[/tex]
u = 3.21 m/s.
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Exercise 1 :On a circuit, a pilot covers 600 m in 7.2 s.1. Calculate its speed in m / s.2. Convert this speed to km / h in two different ways.
Given data
*The distance covers by the pilot is d = 600 m
*The given time is t = 7.2 s
(1)
The formula for the speed is given as
[tex]s=\frac{d}{t}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} s=\frac{600}{7.2} \\ =83.3\text{ m/s} \end{gathered}[/tex]Hence, the speed is s = 83.3 m/s
(2)
The speed converted into km/h as
[tex]\begin{gathered} s=83.3\times(\frac{5}{18}) \\ =299.88\text{ km/h} \end{gathered}[/tex]The second way to convert the speed into kilometer per hour as,
[tex]undefined[/tex]Remember the experiment done by Arthur Holly Compton that demonstrated the particle nature of light (X-rays) definitively. The reaction was:γ+e→γ+e (1) where the outgoing gamma was an X-ray of aa. higher?b. lower?frequency than the initial gamma. Circle your choice.
The correct option is (b)
The outgoing gamma rays are of lower frequency than that of the initial gamma-ray. While investigating the scattering of X-rays, Compton observed that the outgoing rays lose some of their energy in the scattering process and emerge with slightly decreased frequency.
A spring with spring constant 175 N/m has 20 J of EPE stored in it. How much is it compressed?
Given:
• Spring constant = 175 N/m
,• Energy = 20 J
Let's find by how much it is compressed.
Apply the formula:
[tex]E=\frac{1}{2}kx^2[/tex]Where:
E is the energy = 20 J
k is the sring constant = 175 N/m
x is the compression in meters
Rewrite the formula for x:
[tex]x=\sqrt{\frac{2E}{k}}[/tex]Input values and solve for x:
[tex]\begin{gathered} x=\sqrt[]{\frac{2\times20}{175}} \\ \\ x=\sqrt[]{\frac{40}{175}} \\ \\ x=\sqrt[]{0.2285} \\ \\ x=0.48\text{ m} \end{gathered}[/tex]ANSWER:
0.48 m
what is the difference between these two formulas [tex]fc = \frac{m(v {}^{2}) }{r} [/tex][tex]ac = \frac{v {}^{2} }{r} [/tex]and how would I apply them to a promblem?
One is centripetal acceleration (ac) and the other one is centripetal force (fc)
Since = F = m* a
F = force
m= mass
a= acceleration
For centripetal force (Fc), we use centripetal acceleration
Fc= m * ac
ac= centripetal acceleration = v^2 / r
r= radius
v= linear velocity
So, in the end, we have:
Fc = m * v^2 /r
You can apply the to find the Centripetal force of an object with a certain mass, that negotiates a certain radius curve, with a certain speed.
how fast is the angle of depression of the telescope changing when the boat is 190 meters from the shore
ANSWER:
- 0.01943 rad/sec
STEP-BY-STEP EXPLANATION:
The first thing is to make a drawing of what is mentioned in the statement, it would be the following:
Now, we have the following information:
[tex]\begin{gathered} \frac{dy}{dt}=15\text{ m/s} \\ x=50\text{ m} \\ y=190\text{ m} \end{gathered}[/tex]In this right angle triangle formed by telescope of the boat, e can apply the tangent trigonometric ratio, like this:
[tex]\begin{gathered} \tan \theta=\frac{x}{y} \\ \text{ replacing} \\ \theta=\tan ^{-1}\mleft(\frac{x}{y}\mright) \end{gathered}[/tex]Now, we implicitly derive with respect to t:
[tex]\begin{gathered} \frac{d}{dt}(\theta)=\frac{d}{dt}(\tan ^{-1}(\frac{x}{y})) \\ \frac{d}{dt}(\theta)=\frac{1}{1+(\frac{x}{y})^2}\cdot\frac{d}{dt}(\frac{x}{y}) \\ \frac{d}{dt}(\theta)=\frac{y^2}{x^2+y^2}\cdot x\cdot(-\frac{1}{y^2}\cdot\frac{dy}{dt}) \\ \frac{d}{dt}(\theta)=\frac{-x}{x^2+y^2}(\frac{dy}{dt}) \\ \text{ replacing} \\ \frac{d}{dt}(\theta)=\frac{-50}{50^2+190^2}\cdot(15) \\ \frac{d}{dt}(\theta)=-0.01943 \end{gathered}[/tex]The angle of depression is changing at a rate of -0.01943 rad/sec when the boat is 190 m from the shore
what is velocity ratio
Answer:
Definition of velocity ratio
: the ratio of a distance through which any part of a machine moves to that which the driving part moves during the same time.
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Explanation:
assume that the brakes in your car create a constant deceleration if you double your driving speed how does this affect a0 the time required to come to a stop and b) the distance needed to stop?
For the moving car, if you double your driving speed;
a. the time required to come to a stop will increase by a factor of 2.
b. the distance needed to stop will increase by a factor of 2.
What is the formula relating acceleration, speed, time, and distance traveled?The formula relating to the acceleration, speed, time, and distance traveled of an object is a formula of the equations of motion.
The equations of motion are given below as follows:
v = u + at
s = ut + ¹/₂at²
v² = u² + 2as
where:
v = final speed
u = initial speed
t = time taken
s = distance traveled
a = acceleration
Considering the given scenario:
a. if the speed is doubled, the equation v = u + at applies for the time required to come to a stop;
t =2 * v - u / t
the time will be increase.
b. if the speed is doubled, the equation v² = u² + 2as applies for the distance needed to stop;
s = 2 * v² - u²
the distance will increase with an increase in speed.
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How is the direction of the velocity of a satellite differ from the direction of its acceleration?
PLEASE HELP!!!!!
The direction of the velocity of a satellite differ from the direction of its acceleration by an angle of 90° as they act perpendicular to each other.
In an orbital motion, the acceleration of the object is always directed towards the center of the orbit. This acceleration is called as centripetal acceleration. It is denoted by [tex]a_{c}[/tex].
[tex]a_{c}[/tex] = v² / r
In an orbital motion, the velocity of the object is always tangential to the orbit. It can be calculated in two ways,
v = 2 π r / T
v = r ω
Therefore, the direction of the velocity of a satellite differ from the direction of its acceleration by an angle of 90°.
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A turntable turning at rotational speed 78 rpm stops in 45 s when turned off. The turntable's rotational inertia is 1.2 ×10−2kg⋅m2.
Determine the magnitude of the resistive torque that slows the turntable.
The magnitude of the torque that slows the turntable is equal to 2.17 × 10⁻³ N.m.
What is torque?Torque can be described as the measure of the force that can cause the rotation of an object about an axis. The force causes an object to accelerate similarly, torque causes an angular acceleration. Therefore, torque is the rotational equivalent of linear force.
Given, the rotational speed of the turntable, ω = 78 rpm = 8.16 rad/s
The time turntable takes to stop, t = 45 s
The rotational inertia of the turntable, I = 1.2 × 10⁻² Kg.m²
From the rotational kinetics equation, the angular acceleration:
[tex]\omega_f = \omega +\alpha t[/tex]
The final speed of the turntable = 0
[tex]\alpha =\frac{\omega_f-\omega}{t}[/tex]
[tex]\alpha =\frac{0-8.16}{45}[/tex]
[tex]\alpha =-0.1814 \;rad/s^2[/tex]
Calculation of torque by using inertia and angular acceleration:
[tex]\tau = I\alpha[/tex]
[tex]\tau =1.2 \times 10^{-2} \times (-0.1814)[/tex]
[tex]\tau = 2.17\times 10^{-3} \; N.m[/tex]
Therefore, the magnitude of the torque that slows the turntable is 2.17 × 10⁻³ N.m.
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A particle starts from rest of move with constant acceleration 4 m/s² along straight line then, Calculate velocity when it travel distance of 80 m step - by step explaination
Answer:
25.3 m/s
Explanation:
According to the laws of motion, the final velocity and the distance traveled by a particle moving at constant acceleration a m/s² is given by the equation
[tex]v^2 = v_0^2 + 2a(x - x_0)[/tex]
where
v= final velocity m/s
v₀ = initial velocity m/s
x = final position m
x₀ = initial position m
a = acceleration in m/s²
x - x₀ is called the displacement, d
We are given
v₀ = 0 since particle is starting at rest
d = 80 m
a = 4 m/s²
Plugging these values into the above equation.
v² = 0 + 2 x 4 x 80 m²/s²
v² = 640 m²/s²
v = √640 = 8√10 ≈ 25.3 m/s
If sound waves travel at 350 m/s on a warm humid day. And you are submerged underwater. Is the sound you hear while below the surface traveling faster, slower or at the same rate as the sound you hear above the water? Explain your answer
ANSWER:
Sound waves are faster in water
STEP-BY-STEP EXPLANATION:
Sound travels differently in water than it does in air.
Sound waves travel faster in denser substances because neighboring particles will more easily collide with each other.
Water is denser than air, which means that sound waves travel much faster in water than in air.
Although they travel faster in water, we must keep in mind that the human ear evolved to hear sounds in the air and is not as useful when submerged in water.
What is the specific heat c of a 0.500 kg metal sample that rises 5.40 C when 305J of heat is added to it?
ANSWER
[tex]\begin{equation*} 112.96\text{ }J\/kgK \end{equation*}[/tex]EXPLANATION
Parameters given:
Mass of sample, m = 0.5 kg
Temperature change, ΔT = 5.40 °C = 5.40 K
Heat energy, H = 305 J
To find the specific heat capacity of the sample, we have to apply the formula for heat energy:
[tex]H=mc\Delta T[/tex]Where c = specific heat capacity
Therefore, solving for c, the specific heat capacity of the metal sample is:
[tex]\begin{gathered} 305=0.5*c*5.4 \\ c=\frac{305}{0.5*5.4} \\ c=112.96\text{ }J\/kgK \end{gathered}[/tex]That is the answer.
Four charges are arranged in a square formation. Take q to be 1 C of charge and a to be 2 cm in length. Four charges are arranged in a square formation. Find the net electric field at the center of the square.
We have to calculate the electric field in the center, so we need the distance to the center R and the interaction of each charge with that point
[tex]E=\sum_{n\mathop{=}0}^{\infty}E_i[/tex]To calculate the distance R we have a triangle
[tex]R=\sqrt{\frac{2a^2}{4}}=\frac{a}{\sqrt{2}}=0.014m=1.41cm[/tex][tex]\begin{gathered} \sum_{n\mathop{=}0}^{\infty}Ex=\frac{9\cdot10^9}{2\cdot10^{-4}}\cdot(cos45)\cdot(6C)=1.91\cdot10^{14}N/C \\ \sum_{n\mathop{=}0}^{\infty}Ey=4.5\cdot10^{13}\cdot(cos45\degree)(2C)=0.636\cdot10^{14}N/C \\ Etot=\sqrt{Ex^2+Ey^2}=2.01\cdot10^{14}N/C \end{gathered}[/tex]Is important to remember that E has direction so you have to calculate each axis, x, and y
The potential of a 5.0 cm radius conducting sphere is -100 V. What is the charge density on its surface?Group of answer choices-1.8x10-8 C/m23.5x10-8 C/m22.2x10-8 C/m2-2.2x10-8 C/m2-3.5x10-8 C/m2
First lets calculate the surface area
[tex]S=4\cdot\pi\cdot r^2=4\pi\cdot0.05^2m=0.0314m^2[/tex]Now to know the surface charge density we need the next formula:
[tex]CD=\frac{q}{A}[/tex]But we are missing the amount of charge, we only have the potential
So in this case, we going to apply a different formula
[tex]V=\frac{q}{4\cdot\pi\cdot\xi\cdot r}[/tex]q=5.56*10^-10
[tex]CD=-5.56\cdot\frac{10^{-10}}{0.0314}=-1.77\cdot\frac{10^{-8}C}{m^2}[/tex]The anwer might be -1.8x10-8
A flat coil of wire has an area of 0.020 m2 and contains 50 turns. Initially the coil is oriented so that the normal to its surface is parallel to and in the same direction as a constant magnetic field of 0.18 T. The coil is then rotated through an angle of 30o in a time of 0.10 s. What is the average induced emf? -0.44 V +0.44 V +0.24 V
ANSWER:
3rd option: +0.24 V
STEP-BY-STEP EXPLANATION:
Given:
N = 50
Area = 0.020 m^2
B = 0.18 T
θf = 30°
time = 0.10 s
We can calculate the average induced emf by the following formula
[tex]\epsilon=N\cdot B\cdot A\cdot\left(\frac{\cos\theta_i-\cos\theta_f}{t}\right)[/tex]We replacing:
[tex]\begin{gathered} \epsilon=\left(50\right)\left(0.18\right)\left(0.02\right)\left(\frac{\cos\:0\degree\:-\cos\:30\degree}{\:0.1}\right) \\ \epsilon=0.241\cong0.24\text{ V} \end{gathered}[/tex]The correct answer is 0.24V
The cross product of two vectors (X and Y) is a negative vector when the angle between them is:
A. 0
B. 90
C. 180
D. 270
Answer:
C. 180
Explanation:
Answer:
Explanation:
The cross product:
[ a × b ] = | a | · | b | · sin α
sin α < 0 if α ∈ (180°, 360°)
Answer:
D. 270°
A football blocking sled has a mass of roughly 100 kg. If a football player applies a force of 500 N to the sled, and there is a 350 N frictional force acting on the sled, what is the acceleration of the sled?
The the acceleration of the sled of mass 100 Kg will be 1.5 m/s².
What is Friction?A drag is force that opposes the motion of an object by acting in the direction opposite to its motion. It is of two types namely - Static friction and kinetic friction. The static friction is given by - F[S] = μ[s] x η and the kinetic friction is given by F[K] = μ[k] x η.
Given is a football blocking sled which has a mass of roughly 100 kg. A football player applies a force of 500 N to the sled, and there is a 350 N frictional force acting on the sled.
Assume that the force applied by the player is - F[P] = 500 N and the force of friction is - F = 350 N.
Now, for the motion of the sled with acceleration [a], we can write -
F[P] - F = ma
a = {F[P] - F}/m
Substituting the values, we get -
a = (500 - 350)/100
a = 150/100
a = 15/10
a = 1.5 m/s²
Therefore, the the acceleration of the sled of mass 100 Kg will be
1.5 m/s².
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QUESTION 2 (NOVEMBER 2014) Two blocks of masses 20 kg and 5 kg respectively are connected by a light inextensible string, P. A second light inextensible string, Q, attached to the 5 kg block, runs over a light frictionless pulley. A constant horizontal force of 250 N pulls the second string as shown in the diagram below. The magnitudes of the tensions in P and Q are T, and T, respectively. Ignore the effects of air friction.
2.3 Calculate the magnitude of the tension T, in string P. (6)
The magnitude of tension T in string P is 250 N when a constant horizontal force of 250 N pulls the second string.
What is tension and how the tension is calculated out to be 250 N ?Tension is equivalent to pull force , used in most of kinematic questions.Tension can be best explained when you pull a rope or you pull an object during the time period under consideration.Here in this question given, mass of first object is 20 kg , and mass of the second object is 5 kg .Using the equation m1a = T2 - m2a , (m1+m2)a = T2 , (20 +5)250/25 = T2.From this comes the second tension on the string P is 250 N .To know more about tension visit:
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