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4 years ago

15

What force causes a wagon to speed up as it rolls down a hill?

1 answer:

netineya [11]4 years ago

7 0

The answer would be "gravity" because gravity is pulling the wagon down the hill.

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Se pune<br>Numerical problems<br>Convert 100ºC to Kelvin scale

IgorC [24]

Answer:

373 K

Explanation:

To convert from °C to kelvin, you add the value to 273

Therefore,

100°C = (100+273)K

= 373K

5 0

4 years ago

Use the drop-down menu to answer the question. which sound wave-object interaction is used by animals in echolocation?

Dima020 [189]

Answer:

Reflection

Explanation:

6 0

3 years ago

A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if

Zigmanuir [339]

Answer:

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: <em>A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if </em><em>a. </em><em>The mass is doubled? </em><em>b.</em><em> The mass is halved? </em><em>c.</em><em> The amplitude is doubled? </em><em>d.</em><em> The spring constant is doubled? </em>

We have a block-spring system, whose angular frequency ( $\omega$ ) is defined by the following formula:

$\omega = \sqrt{\frac{k}{m} }$ (1)

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

And the period ( $T$ ), measured in seconds, is determined by the following expression:

$T = \frac{2\pi}{\omega}$ (2)

By applying (1) in (2), we get the following formula:

$T = 2\pi\cdot \sqrt{\frac{m}{k} }$

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

8 0

3 years ago

Help please i’ll give brainlist

irga5000 [103]

Answer:

Ans :- C

Explanation:

................

8 0

3 years ago

Read 2 more answers

6. A 15 kg box is given an initial push so that it slides across the floor and comes to a stop.

Illusion [34]

Answer:

(a) Frictional force = 44.145 N

(b) Acceleration = -2.943 m/s²

(c) Distance covered = 1.529 m

Explanation:

Given:

Mass of the box is, $m=15\ kg$

Coefficient of friction is, $\mu =0.30$

Acceleration due to gravity is, $g=9.81\ m/s^2$

Normal force acting on the box is, $N=mg=(15)(9.81) = 147.15\ N$

(a)

Frictional force is given as:

$f=\mu N=0.30\times 147.15=44.145\ N$

Therefore, the frictional force is 44.145 N.

(b)

The acceleration of the box is given using Newton's second law as:

$a=-\frac{f}{m}=-\frac{44.145}{15}=-2.943\ m/s^2$

Therefore, the acceleration of the box is -2.943 m/s².

(c)

Initial velocity is, $u=3.0\ m/s$

Final velocity is, $v=0\ m/s$

Acceleration of the box is, $a=-2.943\ m/s^2$

The displacement of the box can be determined using equation of motion as:

$v^2=u^2+2ad\\0=3^2+2\times (-2.943)d\\0=9-5.886d\\5.886d=9\\d=\frac{9}{5.886}=1.529\ m$

Therefore, the displacement of the box is 1.529 m.

4 0

4 years ago