How much more matter is in a golf ball than a ping pong?

A 0.405 kg mass is attached to a spring with a force constant of 26.3 N/m and released from rest a distance of 3.31 cm from the

miv72 [106K]

Answer:

0.231 m/s

Explanation:

m = mass attached to the spring = 0.405 kg

k = spring constant of spring = 26.3 N/m

x₀ = initial position = 3.31 cm = 0.0331 m

x = final position = (0.5) x₀ = (0.5) (0.0331) = 0.01655 m

v₀ = initial speed = 0 m/s

v = final speed = ?

Using conservation of energy

Initial kinetic energy + initial spring energy = Final kinetic energy + final spring energy

(0.5) m v₀² + (0.5) k x₀² = (0.5) m v² + (0.5) k x²

m v₀² + k x₀² = m v² + k x²

(0.405) (0)² + (26.3) (0.0331)² = (0.405) v² + (26.3) (0.01655)²

v = 0.231 m/s

8 0

4 years ago

A girl playing tug-of-war with her dog pulls the dog a distance of 8.0m by exerting a force at an angle of 18° with the horizont

AnnZ [28]

Answer:

25 N

Explanation:

Work is a product of force and perpendicular distance moved.

W=Fd where F is force exerted and d is perpendicular distance.

However, for this case, the distance is inclined hence resolving it to perpendicular so that it be along x-axis we have distance as $dcos\theta$

Therefore, $W=Fdcos\theta$

Making F the subject of the formula then

$F=\frac {W}{dcos\theta}$ where $\theta$ is the angle of inclination. Substituting 190 J for W then 18 degrees for $\theta$ and 8 m for d then

$F=\frac {190}{8cos18^{\circ}}\approx 25N$

3 0

3 years ago

Josh did an experiment recording the changes in temperature in sand and water when exposed to a light source, and then when the

Marrrta [24]

Before going to solve this question first we have to understand specific heat capacity of a substance .

The specific heat of a substance is defined as amount of heat required to raise the temperature of 1 gram of substance through one degree Celsius. Let us consider a substance whose mass is m.Let Q amount of heat is given to it as a result of which its temperature is raised from T to T'.

Hence specific heat of a substance is calculated as-

$c= \frac{Q}{m[T'-T]}$

Here c is the specific heat capacity.

The substance whose specific heat capacity is more will take more time to be heated up to a certain temperature as compared to a substance having low specific heat which is to be heated up to the same temperature.

As per the question John is experimenting on sand and water.Between sand and water,water has the specific heat 1 cal/gram per degree centigrade which is larger as compared to sand.Hence sand will be heated faster as compared to water.The substance which is heated faster will also cools faster.

From this experiment John concludes that water has more specific heat as compared to sand.

7 0

3 years ago

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7. A block of copper of unknown mass has an initial temperature of 65.4oC. The copper is immersed in a beaker containing 95.7g o

dolphi86 [110]

Answer:

37.34372 kg

Explanation:

m = Mass

$\Delta T$ = Change in temperature

1 denotes water

2 denotes copper

c = Heat capacity

Heat is given by

$Q=mc\Delta T$

In this case the heat transfer will be equal

$m_1c_1\Delta T_1=m_2c_2\Delta T_2\\\Rightarrow m_2=\frac{m_1c_1\Delta T_1}{c_2\Delta T_2}\\\Rightarrow m_2=\frac{95.7\times 4.18(24.2-22.7)}{0.39(65.4-24.2)}\\\Rightarrow m_2=37.34372\ kg$

Mass of copper block is 37.34372 kg

5 0

3 years ago

You drop a rock down a well that is 5.4 m deep. How long does it take the rock to hit the bottom of the well?

Natali5045456 [20]

Equation of motion:

$y_{f}=y_{o}+v_{o}t+ \frac{1}{2} at^{2}$

Since initial velocity is zero, the second term goes away:

$y_{f}=y_{o}+0+ \frac{1}{2} at^{2}$

$y_{f}=y_{o}+\frac{1}{2} at^{2}$

$y_{f}-y_{o}= \frac{1}{2} at^{2}$

$y_{f}-y_{o}=5.4m$

$5.4m= \frac{1}{2} at^{2}$

$\frac{2(5.4m) }{a} = t^{2}$

$a = g = 9.81 \frac{m}{ s^{2}}$

$\frac{2(5.4m) }{9.81 \frac{m}{ s^{2} } } = t^{2}$

$1.1 s^{2} = t^{2}$ $\sqrt{1.1 s^{2}} = \sqrt{t^{2}}$

<u><em> $t = 1.05s$ </em></u>

4 0

3 years ago

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