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Hitman42 [59]
2 years ago
15

The windshield of a speeding car hits a hovering insect. Consider the time interval from just before the car hits the insect to

just after the impact. Which of the following are correct?
A. The system consists of the bug alone.
B. The system consists of the bug plus the car.
C. The magnitudes of the change of velocity are equal for car and bug.
D. The magnitude of change of momentum of the car is bigger that that of the bug.
E. The magnitude of change of momentum of the bug is bigger than that of the car.
F. The system consists of the car alone.
G. The magnitude of change of velocity of the car is bigger than that of the bug.
H. The magnitudes of the change of momentum are equal for car and bug.
I. The magnitude of change of velocity of the bug is bigger than that of the car.
Physics
1 answer:
HACTEHA [7]2 years ago
5 0

Answer:

A. False

B True

C. False

D.False

E. True

F. False

G. False

H. False

I. True

Explanation:

A. False: The system being analyzed consists of the bug and the car.  These are the two bodies involved in the collision.

B. True: The system being analyzed consists of the bug and the car

C. False: The magnitudes of the change in velocity are different from the car and the bug. The velocity of the bug changes from 0 to the velocity of the car, while there is no noticeable change in the velocity of the car

D.False: There is barely any change in the momentum of the car since the mass of the bug is very small.

E. True: Since the mass of the bug is small, and was initially at rest, the magnitude of the change in monentum will be large because the new velocity will be that of the car.

F. False: The system being analyzed consists of the bug and the car. Those are the two bodies involved in the collision

G. False: The car barely changes in velocity since the mass of the bug is small.

H. False: The car barely changes in momentum because the collision does not affect its speed so much. on the other hand the momentum change of the bug is large since its mass is small.

I. True: The bug which was initially at rest will begin moving with the velovity of the speeding car, while the car barely changes in its velocity

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sashaice [31]

Answer:

D

Explanation:

6 0
2 years ago
Current that moves in one direction from negative to positive. May be created by a battery. Is generally NOT found in U.S. Elect
Karolina [17]

Answer:

C. D.C.

Explanation:

The current that is being described here is D.C. or direct current. It is the D.C. that moves in one direction from negative to positive. May be created by a battery. It is different from the A.C.( alternating current whose polarity changes regularly). It is A.C. that is used in electrical outlets and not D.C.So, the current option is C.

3 0
3 years ago
Read 2 more answers
A 20 kg object is dropped from a very tall building. What is the weight of this objects? After 5 seconds, how has the object fal
prohojiy [21]

1. What is the weight of this objects?

Weight is simply the product of mass and gravitational acceleration. Therefore the weight is:

w = 20 kg * 9.81 m/s^2

w = 196.2 kg m/s^2 = 196.2 N

 

2. After 5 seconds, how has the object fallen and what is its speed at this instant?

We can use the formula:

<span>y = v0 t  + 0.5 g t^2</span>

v = v0 + g t

where v0 = 0 since the object starts from rest, y is the distance it fell, t is time

y = 0 + 0.5 * 9.81 * 5^2 = 122.625 m

<span>v = 0 + 9.81 * 5 = 49.05 m/s</span>

8 0
3 years ago
A 0.25-kg ball attached to a string is rotating in a horizontal circle of radius 0.5 m twice every second, what is the tension i
Svetradugi [14.3K]

Answer:

T = 19.75 N

Explanation:

given,

mass of ball = 0.25 Kg

radius = 0.5 m

frequency = 2 s⁻¹

tension in the string = ?

angular velocity

ω = 2 π f

ω = 2 π x 2

ω = 12.57 rad/s

tension on the string is equal to the centripetal force

T = m ω² r

T = 0.25 x 12.57² x 0.5

T = 19.75 N

Tension in the string is equal to T = 19.75 N

3 0
3 years ago
a crude approximation of voice production is to consider the breathing passages and mouth to be a resonating tube closed at one
prohojiy [21]

The fundamental frequency of the tube is 0.240 m long, by taking air temperature to be 37^oC is 367.42 Hz.

A standing wave is basically a superposition of two waves propagating opposite to each other having equal amplitude. This is the propagation in a tube.

The fundamental frequency in the tube is given by

f=\frac{v_T}{4L}

where, v_T=v\sqrt{\frac{T}{273} }

Since, T=37+273 K = 310 K

v = 331 m/s

\therefore v_T=331\sqrt{\frac{310}{273} } = 352.72 \ m/s

Using this, we get:

f=\frac{352.72}{4(0.240)} \\f=367.42 \ Hz

Hence, the fundamental frequency is 367.42 Hz.

To learn more about Attention here:

brainly.com/question/14673613

#SPJ4

7 0
1 year ago
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