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Arada [10]
3 years ago
8

A school bus traveling at 11.1 m/s has a momentum of 152625 kg*m/s what is the mass of the bus

Physics
1 answer:
Juliette [100K]3 years ago
7 0

Answer:

13750 kg

Explanation:

Momentum = mass × velocity

152625 kg m/s = m (11.1 m/s)

m = 13750 kg

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Car A with a mass of 725 kilograms is traveling east at an initial velocity of 15 meters/second. It collides head–on with car B,
ikadub [295]

Answer:

p_t_o_t_a_l=250kg\frac{m}{s}

Explanation:

<u>The total momentum of a system is defined by:</u>

(mv)_t_o_t=m_1v_1+m_2v_2+...

Where,

(mv)_t_o_t is the total momentum or it could be expressed also as p_t_o_t_a_l.

m_1 and m_2 represents the masses of the objects interacting in the system.

v_1 and v_2 are the velocities of the objects of the system.

<em>Remember: </em><em>The momentum is a fundamental physical magnitude of vector type.</em>

We have:

m_1=725 kg

v_1=15\frac{m}{s}\\m_2=625 kg

We are going to take the east side as positive, and the west side as negative. Then the velocity of the car B, has to be <u>negative</u>. It goes in a different direction from car A.

v_2=-17\frac{m}{s}

Then the total momentum of the system is:

p_t_o_t_a_l=m_1v_1+m_2v_2\\p_t_o_t_a_l=(725kg)(15\frac{m}{s})+(625kg)(-17\frac{m}{s})\\p_t_o_t_a_l=10875kg\frac{m}{s}-10625kg\frac{m}{s}\\p_t_o_t_a_l=250kg\frac{m}{s}

8 0
3 years ago
A vehicle travels from a 30m marker to a 100m marker. What is the change in distance?
sesenic [268]
The change in distance is 30 because if you subtract both number you'll get 30
3 0
3 years ago
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Which state of matter is best at conducting heat?
jek_recluse [69]
Solids are the best at conducting heat. 
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3 years ago
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A particle leaves the origin with an initial velocity v → = (3.00iˆ) m/s and a constant acceleration a → = (−1.00iˆ − 0.500jˆ) m
tatiyna

Answer:

the position vector (x,y) will be (1.5 m,-2.25 m) and the velocity vector (vx,vy) will be ( 0 m/s , -1.5 m/s) when x reaches its maximum x coordinate

Explanation:

Since the velocity is related with the acceleration and coordinates through

vx²=v₀x²+2*ax*x

where

vx = velocity in the x direction

v₀x = initial velocity in the x direction = 3 m/s

ax = acceleration in the x direction = −1.00 m/s²

x= coordinates in the x-axis

when x reaches its maximum coordinate , then vx=0

thus

vx²=v₀x²+2*ax*x

0 = (3 m/s)² + 2* (−1.00 m/s²)*x

x= 1.5 m

also for the time t

vx = v₀x + ax*t → t= (vx-v₀x)/ax = (0- 3 m/s)/  (−1.00 m/s²) = 3 seconds

for the y coordinates

y = y₀+v₀y*t + 1/2 ay*t²

where

v₀y = initial velocity in the y direction = 0 m/s

ay = acceleration in the x direction = −0.5 m/s²

y= coordinates in the y-axis

y₀= initial coordinate in the y-axis =0

then since y₀=0 and v₀y=0

y = 1/2*ay*t²

y = 1/2*ay*t² = 1/2*(−0.5 m/s²)*(3 s)² = -2.25 m

and

vy=v₀y+ ay*t= 0+(−0.5 m/s²)*(3 s)= (-1.5 m/s)

therefore the position vector (x,y) will be (1.5 m,-2.25 m)

and the velocity vector (vx,vy) will be ( 0 m/s , -1.5 m/s)

7 0
3 years ago
/Q: DHSMV definition/ i have to write 20 characters for whatever reason lol
cluponka [151]

<span><span>Department of Highway Safety and Motor Vehicles          OR</span></span>

<span><span /><span><span>Division of Highway Safety and Motor Vehicles </span></span></span>

6 0
3 years ago
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