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3 years ago

An ice skater glides for two meters across ice is work done or no work done

Physics

1 answer:

garik1379 [7]3 years ago

5 0

Work is done. work=forcexdisplacement. the ice skater glides 2 meters (displacement), so yes.

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What is the Gravitational Potential Energy of a 30 kg box lifted 1.5 meters off the ground?

juin [17]

Answer:

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

PE = 30 × 9.8 × 1.5

We have the final answer as

Hope this helps you

6 0

3 years ago

The zinc plate is coated with mercury

raketka [301]

Amalgamating is the coating of zinc plate with mercury.

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3 years ago

At a given instant, the force on an electron is in the +z-direction (out of the page), which the electron is moving in the +x-di

Hitman42 [59]

Answer:

The direction of the B-field is in the +y-direction.

Explanation:

The corresponding formula is

$F_B = qv\times B$

This means, we should use right-hand rule.

Our index finger is pointed towards +x-direction (direction of velocity),

our middle finger should point towards the direction of the B-field,

and our thumb should point towards the +z-direction (direction of the force).

Since our middle finger in this situation points towards +y-direction, the B-field should be in +y-direction.

$\^{x} \times \^{y} = \^{z}$

3 0

3 years ago

A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th

NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

$v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2$

$v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)$

$v_{pf}=-31.4\ m/s$

$v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}$

$v_{gf} = 0.0988\ m/s$

4 0

3 years ago

A sealed test tube traps 25.0 cm3 of air at a pressure of 1.00 atm and temperature of 18°C. The test tube’s stopper has a diamet

puteri [66]

Answer:

180° C

Explanation:

First we start by finding the area of the stopper.

A = πd²/4, where d = 1.5 cm = 0.015 m

A = 3.142 * 0.015² * ¼

A = 1.767*10^-4 m²

Next we find the force on the stopper

F = (P - P•)A, where

F = 10 N

P = pressure inside the tube,

P• = 1 atm

10 = (P - 101325) * 1.767*10^-4

P - 101325 = 10/1.767*10^-4

P - 101325 = 56593

P = 56593 + 101325

P = 157918 Pascal

Now, remember, in an ideal gas,

P1V1/T1 = P2V2/T2, where V is constant, then we have

P1/T1 = P2/T2, and when we substitute the values, we have

101325/(273 + 18) = 157918/ T2

101325/291 = 157918/ T2

T2 = (157918 * 291)/101325

T2 = 453 K

T2 = 453 - 273 = 180° C

3 0

4 years ago