A projectile is launched with a velocity of 13.2 m/s at an angle of 37.0° above the horizontal.
What is the speed of the projectile at its highest point?
a. 7.94 m/s
b. 13.2 m/s
c. 10.5 m/s
d. zero

Answer:

The advantages of Hovercraft:

They can travel over almost any non-porous surface.

They can operate to and from any unprepared beach or slipway.

They take fast, direct routes compared to a conventional marine vessel.

Answer:

a.  F = 14,733.33 N

b. F = 221,000 N

Explanation:

Given;

mass of the cars, m = 1300 kg

velocity of the cars, v = 17 m/s

time taken for the first car to stop after hitting a barrier, t = 1.5 s

time taken for the second car to stop after hitting a barrier, t = 0.1 s

The average forces exerted on each car is calculated as follows;

a. the car that hits a line of water barrels and takes 1.5 s to stop

[tex]F = ma = m\times \frac{v}{t} = 1300 \times \frac{17}{1.5} = 14,733.33 \ N\\\\F = 14,733.33 \ N[/tex]

b. the car that hits a concrete barrier and takes 0.10 s to stop

[tex]F = ma = m\times \frac{v}{t}= 1300 \times \frac{17}{0.1} = 221,000 \ N\\\\F = 221,000 \ N[/tex]

Answer:

Explanation:

From the question;

We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.

We are to calculate the following task, i.e. to determine the electric field at the distances:

a)  at 4.75 cm

b)  at 20.5 cm

c) at 125.0 cm

Given that:

the charge (q) = 33.3 nC/m

= 33.3 × 10⁻⁹ c/m

radius of rod = 5.75 cm

a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.

Then, the electric field will be zero.

b) The electric field formula [tex]E = \dfrac{kq }{d}[/tex]

[tex]E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}[/tex]

E = 1461.95 N/C

c) The electric field E is calculated as:

[tex]E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}[/tex]

E = 239.76 N/C

By the work-energy theorem, the total work done on the car is equal to the change in its kinetic energy:

W = ∆K

W = 1/2 (0.34 kg) (22.9 m/s)² - 1/2 (0.34 kg) (6.5 m/s)²

W ≈ 82 J

Answer:

hii

Explanation:

i hope this helps you

Answer:

The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. ... Velocity is a physical vector quantity; both magnitude and direction are needed to define it

Explanation:

hope it helps

pls maek me as brainliest thanks❤

Answer:

1. 0.625 g/mL

2. 2×10⁵ J

3. 5040 J

4. 75000 J

Explanation:

1. Determination of the density.

We'll begin by calculating the volume of the marble. This can be obtained as follow:

Volume of water = 60 mL

Volume of water + Marble = 68 mL

Volume of marble =?

Volume of marble = (Volume of water + Marble) – (Volume of water)

Volume of marble = 68 – 60

Volume of marble = 8 mL

Finally, we shall determine the density of the marble. This can be obtained as follow:

Mass of marble = 5 g

Volume of marble = 8 mL

Density of marble =?

Density = mass / volume

Density of marble = 5 / 8

Density of marble = 0.625 g/mL

2. Determination of the work done.

Mass (m) = 1000 Kg

Height (h) = 20 m

Acceleration due to gravity (g) = 10 m/s²

Workdone (Wd) =?

The work done can be obtained as follow:

Wd = mgh

Wd = 1000 × 10 × 20

Wd= 2×10⁵ J

3. Determination of the potential energy.

Mass (m) = 12 Kg

Height (h) = 42 m

Acceleration due to gravity (g) = 10 m/s²

Potential energy (PE) =?

The potential energy can be obtained as follow:

PE = mgh

PE = 12 × 10 × 42

PE = 5040 J

4. Determination of the kinetic energy.

Mass (m) = 1500 Kg

Velocity (v) = 10 m/s

Kinetic energy (KE) =?

The kinetic energy can be obtained as follow:

KE = ½mv²

KE = ½ × 1500 × 10²

KE = 750 × 100

KE = 75000 J

Answer:

[tex]I_2=0.50 w/m^2[/tex]

Explanation:

From the question we are told that:

initial Intensity [tex]I_1=0.020 w/m^2[/tex]

Final Electric field [tex]E_2=5E[/tex]

Generally the equation for Relation ship between intensity and Electric field is mathematically given by

 [tex]\frac{I_1}{I_2}= \frac{E_1^2}{E_2^2}[/tex]

Therefore

 [tex]I_2=\frac{I_1}{ \frac{E_1^2}{E_2^2}}[/tex]

 [tex]I_2=\frac{0.020}{ \frac{E^2}{5E^2}}[/tex]

 [tex]I_2=0.50 w/m^2[/tex]

Answer:

[tex]d =3.7*10^{-3} m[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=1.2kg[/tex]

Spring constant [tex]\mu=22Nm^{-1}[/tex]

Amplitude [tex]A=5cm=0.05m[/tex]

Generally the equation for displacement d is mathematically given by

 [tex]d = Asin(\omega t)[/tex]

Where

 [tex]\omega=angular\ velocity[/tex]

 [tex]\omega=\sqrt{k/m}[/tex]

 [tex]\omega=\sqrt{22/1.2}[/tex]

 [tex]\omega=4.2817rads^{-1}[/tex]

Therefore  

 [tex]d = 0.05*sin(4.2817*1)[/tex]  

 [tex]d =3.7*10^{-3} m[/tex]

Answer:

a) the required time is 0.6283 μs

b) the inductor current is 0.5 mA

Explanation:

Given the data in the question;

The capacitor voltage has its maximum value of 25 V at t = 0

i.e V[tex]_m[/tex] = V₀ = 25 V

we determine the angular velocity;

ω = 1 / √( LC )

ω = 1 / √( ( 20 × 10⁻³ H ) × ( 8.0 × 10⁻¹² F) )

ω = 1 / √( 1.6 × 10⁻¹³  )

ω = 1 / 0.0000004

ω = 2.5 × 10⁶ s⁻¹

a) How much time does it take until the capacitor is fully discharged for the first time?

V[tex]_m[/tex] =  V₀sin( ωt )

we substitute

25V =  25V × sin( 2.5 × 10⁶ s⁻¹ × t )

25V =  25V × sin( 2.5 × 10⁶ s⁻¹ × t )

divide both sides by 25 V

sin( 2.5 × 10⁶ × t ) = 1

( 2.5 × 10⁶ × t ) = π/2

t = 1.570796 / (2.5 × 10⁶)

t = 0.6283 × 10⁻⁶ s

t = 0.6283 μs

Therefore, the required time is 0.6283 μs

b) What is the inductor current at that time?

[tex]I[/tex](t) = V₀√(C/L) sin(ωt)

{ sin(ωt) = 1 )

[tex]I[/tex](t) = V₀√(C/L)

we substitute

[tex]I[/tex](t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )

[tex]I[/tex](t) = 25 × 0.00002

[tex]I[/tex](t) = 0.0005 A

[tex]I[/tex](t) = 0.5 mA

Therefore, the inductor current is 0.5 mA

The time taken for the capacitor to fully discharge is 6.28 x 10⁻⁷ s.

The current in the inductor at the given time is 0.0005 A.

The angular velocity of the circuit is calculated as follows;

[tex]\omega = \frac{1}{\sqrt{LC} } \\\\\omega = \frac{1}{\sqrt{20 \times 10^{-3} \times 8 \times 10^{-12} } } \\\\\omega = 2.5 \times 10^6 \ rad/s[/tex]

V = V₀(sinωt)

sinωt = V/V₀

sinωt = = 25/25

sin(ωt) = 1

ωt = sin⁻¹ (1)

ωt = π/2

2.5 x 10⁶ t = π/2

t = 1.57 / (2.5 × 10⁶)

t = 6.28 x 10⁻⁷ s

The current in the inductor at the given time is calculated as follows

[tex]I(t) = V_0 \sqrt{\frac{C}{L} } \\\\I = 25 \times \sqrt{\frac{8\times 10^{-12}}{20 \times 10^{-3}} } \\\\I = 0.0005 \ A[/tex]

Learn more about inductor current here: https://brainly.com/question/4425414

Answer:

NORMAL

Explanation:

In mathematics to define a plane you need a line perpendicular to the plane called NORMAL and the point of application of this line at a point on the plane.

Based on this definition you only need to specify the normal plane is perpendicular to this line, it should be noted that this does not define a single plane but the whole family of plane is contains the normal.

Consequently only the NORMAL is needed

Answer:

Tx not but mybe

Explanation:

for that reason its just trying to help

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore,  the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F =  624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

Answer:

Here are a few commonly used types of shooting in basketball.

Answer:

the period of oscillation of the given object is 0.14 s

Explanation:

Given;

mass of the object, m = 3 kg

extension of the spring, x = 0.085 m

The spring constant is calculated as follows;

[tex]F = mg = \frac{1}{2} ke^2\\\\2mg = ke^2\\\\k = \frac{2mg}{e^2} \\\\k = \frac{2\times 3 \times 9.8}{(0.085)^2} \\\\k = 8,138.41 \ N/m[/tex]

The angular speed of a 4 kg object is calculated as follows;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\frac{2\pi }{T} = \sqrt{\frac{k}{m} } \\\\T= 2\pi \sqrt{\frac{m}{k} } \\\\T = 2\pi \sqrt{\frac{4}{8138.41} }\\\\T = 0.14 \ s[/tex]

Therefore, the period of oscillation of the given object is 0.14 s

Answer:

Explanation:

Reverse the second equation, and change the sign of the enthalpy

This question is incomplete, the complete question is;

In June 1985, a laser beam was sent out from the Air Force Optical Station on Maui and reflected back from the shuttle Discovery as it passed by 354 km overhead. The diameter of the central maximum of the beam at the shuttle position was said to be 9.1 m, and the beam wavelength was 500 nm.

What is the effective diameter of the circular laser aperture at the Maui ground station

Answer:

the effective diameter of the circular laser aperture at the Maui ground station is 4.747 cm

Explanation:

Given the data in the question;

Separation between observer and point L = 354 km = 354000 m

Linear separation D = 9.1 m

wavelength λ = 500 nm = 500 × 10⁻⁹ m

Now, for small angles;

θ = D / L

θ = 9.1 m /  354000 m

θ = 2.57 × 10⁻⁵ rad

For a circular aperture;

sinθ = ( 1.22 × λ ) / d

for small angles;

θ = ( 1.22 × λ ) / d

so

θ = 2 × θ

θ = 2 × [( 1.22 × λ ) / d]

we substitute

2.57 × 10⁻⁵ = 2 × [( 1.22 × 500 × 10⁻⁹ ) / d]

2.57 × 10⁻⁵ = 0.00000122 / d

d = 0.00000122 / 2.57 × 10⁻⁵

d = 0.04747 m

d = ( 0.04747 × 100 )m

d = 4.747 cm

Therefore, the effective diameter of the circular laser aperture at the Maui ground station is 4.747 cm

Answer:

There are a couple of interesting things about kinetic energy that we can see from the equation.

Kinetic energy depends on the velocity of the object squared. This means that when the velocity of an object doubles, its kinetic energy quadruples. A car traveling at 60 mph has four times the kinetic energy of an identical car traveling at 30 mph, and hence the potential for four times more death and destruction in the event of a crash.

Kinetic energy must always be either zero or a positive value. While velocity can have a positive or negative value, velocity squared is always positive.

Kinetic energy is not a vector. So a tennis ball thrown to the right with a velocity of 5 m/s, has the exact same kinetic energy as a tennis ball thrown down with a velocity of 5 m

Answer:

[tex]t=2413s[/tex]

Explanation:

From the question we are told that:

Velocity [tex]v=1.5m/s[/tex]

Time [tex]t=30min=>30*60=>1800[/tex]

Distance [tex]d=24.3km[/tex]

Generally the Newton's equation for Speed going down the stream is mathematically given by

 [tex]v + u = \frac{d}{t}[/tex]

 [tex]1.5+v=frac{24300}{1800}[/tex]

 [tex]v=12m/s[/tex]

Therefore

 [tex]v + u = \frac{d}{t}[/tex]

 [tex]t=\frac{24300}{12-1.5}[/tex]

 [tex]t=2413s[/tex]

Answer:

The second one says cant open file............

Answer:

F₂ = -7.3 N

Explanation:

Given that,

The mass of an object, m₁ = 3.7 kg

First force, F₁ = 11 N

The net acceleration of the object is 1 m/s².

We know that,

F₁+F₂ = ma

11+F₂ = (3.7)(1)

F₂ = 3.7-11

F₂ = -7.3 N

so, the other force is 7.3 N and it is acting in west direction.

Answer:

A charge is a physical entity that has been quantized. The limits on its values are the value of a charged particle quantized in the state where 'n'...

Explanation:

One example of a physical entity that is quantized is:

The amount of money in your pocket.

The amount can't have any fraction of 1 cent.  

Its value must be an integer-multiple of cents, or 0.01 dollar.    

When it increases or decreases, it jumps from one integer number of cents to the next integer number.  It doesn't "slide" from one to the next.  It can never have a value between two integer numbers of cents.

Answer:

D. digital

Explanation:

Digital signals are transmitted through electromagnetic waves.

I'm not sure

Answer:

Because molecules are to small for us people to see with the bare eye, unless you use a telescope.

Explanation:

Explanation:

Because in solid form, the molecules stood in place, not moving. In liquid state, the molecules move slowly but not as fast as air molecules.

Answer:

0 N

Explanation:

Applying,

F = qvBsin∅................. Equation 1

Where F = Force on the charge, q = charge, v = Velocity, B = magnetic charge, ∅ = angle between the velocity and the magnetic field.

From the question,

Given: q = 4.88×10⁻⁶ C, v = 265 m/s, B = 0.0579 T, ∅ = 0°

Substitute these values into equation 1

F = ( 4.88×10⁻⁶)(265)(0.0579)(sin0)

Since sin0° = 0,

Therefore,

F = 0 N

Answer:

The maximum speed is 18.86 m/s.

Explanation:

initial  radius, r = 150 m

maximum speed, v = 26.5 m/s

new radius, r' = 76 m

Let the new maximum speed is v'.

The formula of the maximum speed is

[tex]tan\theta = \frac{v^2}{rg}[/tex]

So,

[tex]\frac{v'^2}{v^2}=\frac{r'}{r}\\\\\frac{v'^2}{26.5\times 26.5}=\frac{76}{150}\\\\v=18.86 m/s[/tex]

Answer:

The resistance of A is 6 ohms and the resistance of B is 3 ohms

Explanation:

Step 1: For the first connection (parallel connection), the resistance of B will be calculated.

Note: in a parallel connection, the voltage through each resistor is the same.

[tex]V = I_AR_A = I_BR_B\\\\R_B = \frac{V}{I_B} = \frac{6}{2} = 3 \ ohms[/tex]

Step 2: The resistance of A will be calculated from the second connection (series connection)

Note: in series connection, the current flowing in each resistor is the same

[tex]V = V_A + V_B\\\\V = IR_A + IR_B\\\\The \ voltage \ drop \ in \ B; \ V_B = V- V_A\\\\V_B = 6 - 4 = 2 \ V\\\\IR_B = 2\ V\\\\I = \frac{2 \ V}{R_B}= \frac{2}{3} \ A\\\\The \ resistance \ of \ A \ is \ calculated \ as ;\\\\IR_A = 4 \ V\\\\R_A = \frac{4}{I} = \frac{4 \times 3}{2} = 6 \ ohms[/tex]

Answer:

The maximum magnetic force is 2.637 x 10⁻¹² N

Explanation:

Given;

Power, P = 8.25 m W = 8.25 x 10⁻³ W

charge of the radiation, Q = 1.12 nC = 1.12 x 10⁻⁹ C

speed of the charge, v = 314 m/s

area of the conecntration, A = 1.23 mm² = 1.23 x 10⁻⁶ m²

The intensity of the radiation is calculated as;

[tex]I = \frac{P}{A} \\\\I = \frac{8.25 \times 10^{-3} \ W}{1.23 \ \times 10^{-6} \ m^2} \\\\I = 6,707.32 \ W/m^2[/tex]

The maximum magnetic field is calculated using the following intensity formula;

[tex]I = \frac{cB_0^2}{2\mu_0} \\\\B_0 = \sqrt{\frac{2\mu_0 I}{c} } \\\\where;\\\\c \ is \ speed \ of \ light\\\\\mu_0 \ is \ permeability \ of \ free \ space\\\\B_0 \ is \ the \ maximum \ magnetic \ field\\\\B_0 = \sqrt{\frac{2 \times 4\pi \times 10^{-7} \times 6,707.32 }{3\times 10^8} } \\\\B_0 = 7.497 \times 10^{-6} \ T[/tex]

The maximum magnetic force is calculated as;

F₀ = qvB₀

F₀ = (1.12 x 10⁻⁹) x (314) x (7.497 x 10⁻⁶)

F₀ = 2.637 x 10⁻¹² N