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Vsevolod [243]
3 years ago
8

Ou are driving on the highway at a speed of 40 m/s (which is over the speed limit) when you notice a cop in front of you. To avo

id a ticket, you press on the brake and slow to a speed of 30 m/s over the course of 5 seconds. What is the acceleration of the car? SHOW WORK OR GET REPORTED(have 5 accounts)
What is your car's initial velocity?

What is your car's final velocity?

How long does it take the car to slow down?

Write the equation you will use to solve this problem.

What is the acceleration of your vehicle?
+ 2.0 m/s^2
- 2.0 m/s^2
+ 8.0 m/s^2
- 6.0 m/s^2
Physics
1 answer:
harina [27]3 years ago
7 0

Answer:

What is the acceleration of your vehicle?

+ 2.0 m/s^2✔

- 2.0 m/s^2

+ 8.0 m/s^2

- 6.0 m/s^2

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The distance between two planets is 1600 km. How much time would the light
Snowcat [4.5K]

Answer:

5.33*10^-3 seconds

Explanation:

c = d/t

c = speed of light constant (3.0*10^5 km/s)

d = distance (1600 km)

t = ?

3.0*10^5 = 1600/t

t = 1600/3.0*10^5

t = 5.33*10^-3 seconds

I hope this helped! :)

6 0
3 years ago
Two factor that affect gravitational force​
Gnoma [55]
When dealing with the force of gravity between two objects, there are only two things that are important – mass, and distance. The force of gravity depends directly upon the masses of the two objects, and inversely on the square of the distance between them.
6 0
3 years ago
3
Misha Larkins [42]

Hi there!

The maximum deformation of the bumper will occur when the car is temporarily at rest after the collision. We can use the work-energy theorem to solve.

Initially, we only have kinetic energy:

KE = \frac{1}{2}mv^2

KE = Kinetic Energy (J)
m = mass (1060 kg)
v = velocity (14.6 m/s)

Once the car is at rest and the bumper is deformed to the maximum, we only have spring-potential energy:

U_s = \frac{1}{2}kx^2

k = Spring Constant (1.14 × 10⁷ N/m)

x = compressed distance of bumper (? m)

Since energy is conserved:

E_I = E_f\\\\KE = U_s\\\\\frac{1}{2}mv^2 = \frac{1}{2}kx^2

We can simplify and solve for 'x'.

mv^2 = kx^2\\\\x = \sqrt{\frac{mv^2}{k}}

Plug in the givens and solve.

x = \sqrt{\frac{(1060)(14.6^2)}{(1.14*10^7)}} = \boxed{0.0198 m}

3 0
2 years ago
A motorcycle together with its Rider with 150 kg how much work is needed to set its movement at 15 metre per second.​
AVprozaik [17]

Answer:

W = 16875 J

Explanation:

We have,

Mass of a motorcycle together with its rider is 150 kg

It is required to find the work is needed to set its movement at 15 metre per second. The work done is given by :

W=\dfrac{1}{2}mv^2

Plugging all the values in above formula such that :

W=\dfrac{1}{2}\times 150\times (15)^2\\\\W=16875\ J

So, the work of 16875 J is needed.

5 0
3 years ago
on takeoff, the propellers on a uav (unmanned aerial vehicle) increase their angular velocity from rest at a rate of ω = (25.0t)
Rudik [331]

Hi there!

a)
We can find the angular velocity at t = 2.0 s by plugging in this value into the equation.

\omega (t) = 25.0t \\\\\omega (2) = 25.0(2) = \boxed{50.0 \frac{rad}{s}}

b)

The angular acceleration is the derivative of the angular velocity, so:
\alpha (t) = \frac{d\omega}{dt} (25.0t) = 25.0

Thus, the angular acceleration is a <u>constant 25 rad/s².</u>

7 0
2 years ago
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