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Shkiper50 [21]
3 years ago
10

A motorcycle has a speed of 30 m / s. After braking, it decelerates with constant deceleration A = -3.0 m / s ^ 2

Physics
2 answers:
kenny6666 [7]3 years ago
7 0

Answer:21

Explanation:

marin [14]3 years ago
4 0

a = \displaystyle\frac{v-u}{t}

-3 = \displaystyle\frac{v - 30}{3}

v -30 = -9

v = 21 m/s

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A bullet of .05 kg is fired into a block of wood. Knowing that the bullet left the gun with a muzzle velocity of 350. m/s, and t
Lunna [17]

We know, Impulse of a body is change in momentum.

Also, the final velocity of wood block is 0 m/s.

So, Impulse = Final Momentum - Initial Momentum

Impulse = 0 - ( 0.05 kg × 350 m/s )

Impulse = - 17.5 kg m/s

Therefore, the impulse exerted by the wood on the bullet is -17.5 kg m/s .

6 0
3 years ago
A car of mass 1500 kg travels due East with a constant speed of 25.0 m/s. Eventually it turns right, and travels due South with
e-lub [12.9K]

Answer:

The direction of the car’s change in linear momentum is 149.04° West of North

Explanation:

Momentum is defined as the product of mass of a body and its velocity

Momentum = mass × velocity

Change in Momentum = mass × change in velocity

∆P = m∆v

∆P = m(v-u)

Given m = 1500kg

v = 25m/s

u = 15m/s

∆P = 1500(25-15)

∆P = 1500×10

∆P = 15,000kgm/s

Since the car first travels due East i.e +x direction

x = 25m/s

Travelling due south is negative y direction

y = -15m/s

Direction of the car change

θ = tan^-1(y/x)

θ = tan^-1(-15/25)

θ = tan^-1(-0.6)

θ = -30.96°

Since tan is negative in the second quadrant

θ = 180-30.96

θ = 149.04°

The direction of the car’s change in linear momentum is 149.04° West of North

5 0
4 years ago
7. A Cyclist speeds up from 6.00 m/s to 9.70 m/s in a time of 1.40 seconds.
disa [49]

Answer:

2.64m/s

Explanation:

9.7m/s - 6m/s = 3.7

3.7 / 1.7 = 2.64

7 0
3 years ago
Read 2 more answers
Column A
Nikolay [14]

Answer:

74747463476869+7t87589598

Explanation:

thats your answer

6 0
3 years ago
Calculate the root mean square velocity of the nitrogen molecules, in meters per second (m/s). Do not include the unit of measur
kipiarov [429]

This question involves the concepts of average kinetic energy and velocity.

Complete question: Calculate the root mean square velocity of the nitrogen molecules, at 25°C, in meters per second (m/s). Do not include the unit of measure in your answer.

The root mean square velocity of the nitrogen molecule at 25°C is "727.66 m/s".

The average kinetic energy of gas molecules is given by the following formula:

K.E=\frac{3}{2}KT

Also,

K.E = \frac{1}{2}mv^2

Comparing both equations we get:

\frac{3}{2}KT=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{3KT}{m}}

where,

v = root men square velocity = ?

K = Boltzman's constant = 1.38 x 10⁻²³ J/k

T = absolute temperature = 25°C + 273 = 298 k

m = mass of nitrogen molecule = 2.33 x 10⁻²⁶ kg

Therefore,

v=\sqrt{\frac{3(1.38\ x\ 10^{-23}\ J/k)(298\ k)}{2.33\ x\ 10^{-26}\ kg}}

<u>v = 727.66 m/s</u>

<u></u>

Learn more about the root mean square velocity here:

brainly.com/question/2020688?referrer=searchResults

6 0
3 years ago
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