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aleksandrvk [35]
3 years ago
6

Two cars are traveling on a desert road. After 5.0 seconds they are side by side at the next telephone pole.The distance between

the poles is 112m.
Identify the following quantities:
a. the displacement of car A after 5.0s
b. the displacement of car B after 5.0s
 c. the average velocity of car A during 5.0s
d. the average velocity of car B during 5.0s
Physics
1 answer:
stira [4]3 years ago
8 0
The question looks incomplete, but according to the information given above seem like they have <span>identical journeys.
</span>a. the displacement of car A - <span> 65.5 m 
</span>b. the displacement of car B  - <span>65.5 m
c. average velocir</span>y of A  65.5 / 5 = 13.1 m/s
d. the average velocity of car B  has the same. 
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A Carnot engine whose high-temperature reservoir is at 464 K has an efficiency of 25.0%. By how much should the temperature of t
nexus9112 [7]

Answer

given,

high temperature reservoir (T_c)= 464 K

efficiency  of reservoir (ε)= 25 %

temperature to decrease = ?

increase in efficiency = 42 %

now, using equation

 \epsilon = 1 - \dfrac{T_C}{T_H}

 0.25 = 1 - \dfrac{T_C}{464}

 T_C= (1 - 0.25) \times 464

 T_C= 0.75 \times 464

      T_C = 348 K

now,

if the efficiency is equal to 42$ = 0.42

 T_C= (1 - 0.42) \times 464

 T_C= 0.58 \times 464

 T_C= 269.12\ K

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3 years ago
What is the study of energy and how it affects matter called?
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What variable does each of the symbols in the equation represent?
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2 years ago
An airplane has a momentum of 8.55 x 107 kg.m/s[S] and a velocity of 900 km/h[S]. Determine the mass of the airplane.
kow [346]

Answer:

342,000kg

Explanation:

p=mv

8.55*10^7 kg*m/s=m(900 km/h)

85,500,000 kg*m/s=m(900 km/h)

(85,500,000 kg*m/s)/(900 km/h)=m

Get same units.... 900km/h = 250m/s

m/s cancel in the division, you are left with just kg!!

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6 0
2 years ago
Two people stand facing each other at roller skating rink then push off each other
9966 [12]

a) 0 kg m/s

b) 0 kg m/s

c) +3 m/s

d) 60 N

Explanation:

a)

The momentum of an object is a vector quantity given by:

p=mv

where

m is the mass of the object

v is the velocity of the object

In this problem, we have a system of two people, so the total momentum will be the sum of the individual momenta of the two people:

p=p_1 + p_2

Which can be rewritten as

p=m_1 u_1 + m_2 u_2

where m_1,m_2 are the masses of the two people and u_1,u_2 their initial velocities.

However, the two people are initially at rest, so

u_1 = 0\\u_2 = 0

Therefore the total momentum is

p=0+0=0

b)

The principle of conservation of momentum states that when there are no external forces acting on a system, the total momentum of the system is conserved, so we can write:

p_i = p_f

where

p_i is the total momentum of the system before

p_f is the total momentum of the system after

In this problem,

p_i = 0

As we calculated in part a: this is because the total momentum of the two people before they push off each other is zero.

Therefore, according to the law of conservation of momentum,

p_f = p_i = 0

So the total momentum is zero also after they push off each other.

c)

The total momentum of the girl and the boy after they push off each other can be written as:

p_f = m_1 v_1 + m_2 v_2 (1)

where:

m_1 = 30 kg is the mass of the girl

v_1 = -5 m/s is her velocity (she moves backward, so the negative sign)

m_2 = 50 kg is the mass of the boy

v_2 is the velocity of the boy

As calculated in part b), we also know that the total momentum of the girl and the boy is

p_f = 0 (2)

By combining eq(1) and eq(2) we get

0=m_1 v_1 + m_2 v_2

And solving for v2 we find the velocity of the boy:

v_2=-\frac{m_1 v_1}{m_2}=-\frac{(30)(-5)}{50}=+3 m/s

and the positive sign means he is moving forward.

d)

We can solve this part by applying the impulse theorem, which states that the change in momentum of an object is equal to the product between the force applied on it and the duration of the collision:

\Delta p = F\Delta t

where

\Delta p is the change in momentum

F is the force

\Delta t is the time during which the force is applied

In this problem:

\Delta t = 2.5 s

For the boy, the change in momentum is:

\Delta p = m_2 (v_2 - u_2)

And since

m_2 = 50 kg\\u_2 = 0 m/s\\v_2 = 3 m/s

We have

\Delta p = (50)(3-0)=150 kg m/s

So, the force exerted between the boy and the girl is:

F=\frac{\Delta p}{\Delta t}=\frac{150}{2.5}=60 N

8 0
3 years ago
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