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Dahasolnce [82]
2 years ago
4

A car (of mass 1500 kg) and a small truck (of mass 2000 kg) collide suddenly at right angles at an icy intersection. The car was

originally traveling eastward at 20.0 m/s and the truck was traveling northward at 20.0 m/s just before the collision took place. What is the speed of the combined wreck just after the collision if the vehicles become enmeshed?
Physics
1 answer:
SVEN [57.7K]2 years ago
5 0

Answer:

The answer is 20.0 m/s. Let's explain why.

Explanation:

In the crash problems, the key is that the<u> total moment is the same </u>before and after the crash

P_{beforecrash}=P_{aftercrash}

and P is defined as P=mV

Let's say the car has m_{1} mass and V_{1} speed and the small truck has m_{2} mass and V_{2} speed.

Total moment before the crash would be:

m_{1} V_{1} +m_{2} V_{2}

And after the crash the it equals (m_{1} +m_{2})V_{combined}

As we said before total moment is equal before and after the crash.

m_{1} V_{1} +m_{2} V_{2}=(m_{1} +m_{2})V_{combined}

1500.20+2000.20=(1500+2000)V_{combined}

Therefore

V_{combined}=20

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A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a very light rope that goes over an ideal pulley. If the
AleksAgata [21]

Answer:

acceleration = 2.4525‬ m/s²

Explanation:

Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²

Tension in the rope = T

Sol: m2 > m1

i) for downward motion of m2:

m2 a = m2 g - T

5 a = 5 × 9.81 m/s² - T  

⇒ T = 49.05‬ m/s² - 5 a     Eqn (a)‬

ii) for upward motion of m1

m a = T - m1 g

3 a = T - 3 × 9.8 m/s²

⇒ T =  3 a + 29.43‬ m/s²   Eqn (b)

Equating Eqn (a) and(b)

49.05‬ m/s² - 5 a = T =  3 a + 29.43‬ m/s²

49.05‬ m/s² - 29.43‬ m/s² = 3 a + 5 a

19.62 m/s² = 8 a

⇒ a = 2.4525‬ m/s²

5 0
2 years ago
An electric drill transfers 200 J of energy into a useful kinetic energy store. It also transfers 44 J of energy by sound and 48
kramer

Answer:

Er = 108 [J]

Explanation:

To solve this problem we must understand that the total energy is 200 [J]. Of this energy 44 [J] are lost in sound and 48 [J] are lost in heat. In such a way that these energy values must be subtracted from the total of the kinetic energy.

200 - 44 - 48 = Er

Where:

Er = remaining energy [J]

Er = 108 [J]

3 0
2 years ago
A particular string resonates in four loops at a frequency of 320 Hz . Name at least three other (smaller) frequencies at which
goldfiish [28.3K]

Answer:

160 Hz  ,  240 Hz  , 400 Hz

Explanation:

Given that

Frequency of forth harmonic is 320 Hz.

Lets take fundamental frequency = f₁

f_1=\dfrac{320}{4}\ Hz

f₁=80 Hz

Frequency of first harmonic = f₂

f₂=2 f₁

f₂ =2 x 80 = 160 Hz

Frequency of second harmonic = f₃

f₃= 3 f₁=3 x 80 = 240 Hz

Frequency of fifth harmonic = f₅

f₅=  5 f₁= 5 x 80 = 400 Hz

Three frequencies are as follows

160 Hz  ,  240 Hz  , 400 Hz

6 0
2 years ago
Fill in the blanks to complete the statements.
Murljashka [212]

Answer:

When an object changes speed (increases/decreases) it results in acceleration/de acceleration, its velocity also changes.

Explanation:

Acceleration is the rate of change in velocity.An object can accelerate when speed increases, decreases or direction changes. All these instances involves a change in velocity.Velocity is a vector quantity thus it has magnitude and the direction.Acceleration due to change in direction is centripetal acceleration.The expression for finding acceleration is;

a=change in velocity/change in time

a=Δv/Δt in m/s²

3 0
2 years ago
A moving roller coaster speeds up with constant acceleration for 2.3\,\text{s}2.3s2, point, 3, start text, s, end text until it
Ad libitum [116K]

Answer:

Δx=(v+v0/2)t

Explanation:

We can figure out which kinematic formula to use by choosing the formula that includes the known variables, plus the target unknown.

In this problem, the target unknown is the initial velocity v_0v  

0

​  

v, start subscript, 0, end subscript of the roller coaster.

7 0
2 years ago
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