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2 years ago

Social loafing is most likely to occur _____.

Physics

1 answer:

loris [4]2 years ago

5 0

Answer:

Answer is when no single person in a group is being watched.

Refer below.

Explanation:

Social loafing is most likely to occur:

when no single person in a group is being watched.

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If a change is made to a system in equilibrium, the equilibrium will shift to oppose the change.

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Explanation:

Because it won’t

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A student pulls a cart across the floor with a force of 45 N. If the cart travels 10 m, how much work does the cart do on the st

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I'm assuming we're applying the standard Integral form of the calculation of work. The solution is provided in the image.

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2 years ago

How much work does gravity do on the compressor during this displacement?

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Explanation and Examples

let the mass of the compressor be

mass (m):

height in x axis is (h1)

height in y axis be (h2):

Height difference: h2-h1

displacement x force:

mass x gravity x height

(m)*9.8*(height difference) = ___ J

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a soccer player kicks a ball. According to Newton's third law of motion, the player's foot exerts a force on the ball. The ball

Leto [7]

Answer:

The forces are exerted on different objects so they are not balanced forces.

Explanation:

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2 years ago

A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas

Radda [10]

Answer:

change in momentum, $\Delta p=7.65 \,kg.m.s^{-1}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

Average Force, $F=144.3396\,N$

$F_x=125.0018\,N$

$F_y=72.1698\,N$

Explanation:

Given:

angle of kicking from the horizon, $\theta= 30^{\circ}$

velocity of the ball after being kicked, $v=18 m.s^{-1}$

mass of the ball, $m=0.425\, kg$

time of application of force, $t=5.3\times 10^{-2}\,s$

We know, since body is starting from the rest

$\Delta p=m.v$ .....................(1)

$\Delta p=0.425\times 18$

$\Delta p=7.65 \,kg.m.s^{-1}$

Now the components:

$\Delta p_x= 7.65\times cos 30^{\circ}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

similarly

$\Delta p_y= 7.65\times sin 30^{\circ}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

also, impulse

$I=F\times t$ .........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

$F\times t=m.v$

$F\times 5.3\times 10^{-2}= 7.65$

$F=144.3396\,N$

Now, the components

$F_x=144.3396\times cos 30^{\circ}$

$F_x=125.0018\,N$

$F_y=144.3396\times sin 30^{\circ}$

$F_y=72.1698\,N$

6 0

2 years ago