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3 years ago

A rubber ball moving at a speed of 5 m/s hit a flat wall and returned to the thrower at 5 m/s. Which statement

Physics

1 answer:

Step2247 [10]3 years ago

4 0

Answer:

d. Its magnitude and its direction both remained the same.

Explanation:

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

$Momentum = mass * velocity$

The law of conservation of momentum states that the total linear momentum of any closed system would always remain constant with respect to time.

This ultimately implies that, the law of conservation of momentum states that if objects exert forces only on each other, their total momentum is conserved.

In this scenario, a rubber ball moving at a speed of 5 m/s hit a flat wall and returned to the thrower at 5 m/s. Thus, the statement which correctly describes the momentum of the rubber ball is that its magnitude and its direction both remained the same because its velocity didn't change while returning to the thrower.

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Shadow is formed when an_____<br>object comes in the<br>way<br>of light.

Zanzabum

Answer:

shadow is formed when an opaque object comes on path of light shadows are formed because light travels in a straight line and it can't pass through an opaque object

7 0

3 years ago

Read 2 more answers

A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

5 0

3 years ago

A sailcraft is stalled on a windless day. A fan is attached to the craft and blows air into the sail which bounces backward upon

mylen [45]

Answer:

Impulse = change in momentum w bounce

There are 2 impulses acting. Recoil of the fan going the negative direction and the impulse of the air bouncing off the sail. The greater impulse will bounce so the direction will be to the right moving the craft.

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2 years ago

A block is sent up a frictionless ramp along which an x axis extends upward. The figure below gives the kinetic energy of the bl

ss7ja [257]

Kinetic energy =1/2 mv^2

<span>m=2ke/v^2 </span>

<span>m=2(34)/3.6^2 </span>

<span>m=5.24 </span>

<span>force normal = mg </span>
<span>=5.24 x 9.8 </span>
<span>force normal = 51.4N

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

</span>

5 0

3 years ago

A person stands 6.00 m from a speaker, and 8.00 m from an identical speaker. What is the wavelength of the first (n=1) interfere

dalvyx [7]

Answer:

wavelength = 4 m

Explanation:

For distance 6 and 8m and speed of sound in air = c.

The travel time form the various distances 6 and 8 are 6/c and 8/c respectively.

cos(wt1) + cos(wt2) = 0

for a shift in phase t1 = t - 6/c,

t2 = t - 8/c

substituting t1 and t2

cos(π - w(t - 8/c)) = cos(w(t - 6/c))

solving using trigonometry identities in radians.

we have,

π - 2πn = w(t - 8/c) - w(t - 6/c)

putting w = 2πf

π - 2πn = 2πf(t - 8/c) - 2πf(t - 6/c)

dividing both sides by π

1 - 2n = 2ft - 16(f/c) - 2ft + 12(f/c)

simplifying we have,

1 - 2n = -4(f/c)

solving for f we have,

f = c/4(2n - 1)

putting n=1 and c = 343m/s

f = (343/4)*(2(1) - 1)

f = 85.75 Hertz

wave lenght = c/f , where c= speed of sound in air , f= frequency

wave lenght = 343/85.75 = 4m

5 0

3 years ago