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2 years ago

A force of 6.0 N gives a 2.0 kg block an acceleration of 3.0

Physics

1 answer:

Semenov [28]2 years ago

7 0

Explanation:

The Net Force of the object can be written by:

Fnet = ma

where m is the mass of the object in kg

a is the acceleration of the object in m/s^2

Hence by applying the formula we get:

Fnet = (2.0)(3.0)

= 6N

We also know that Net force is also the sum of all forces acting on an object. In this case Friction and the Pushing Force is acting on the object. Hence we can write that:

Fnet = Pushing Force + (-Friction)

6N = 6N - Friction

Friction = 0N

Hence the force of friction is 0N.

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A sprinter runs 500 meters west in a straight line and then turns around and runs in the same

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Answer:

200 meters

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3 years ago

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Answer:

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3 years ago

The kinetic energy of an object with a mass of 6.8 kg and a velocity of 5.0 m/s is J. (Report the answer to two significant figu

abruzzese [7]

Answer:

$\boxed{\sf Kinetic \ energy \ (KE) = 85 \ J}$

Given:

Mass (m) = 6.8 kg

Speed (v) = 5.0 m/s

To Find:

Kinetic energy (KE)

Explanation:

Formula:

$\boxed{ \bold{\sf KE = \frac{1}{2} m {v}^{2} }}$

Substituting values of m & v in the equation:

$\sf \implies KE = \frac{1}{2} \times 6.8 \times {5}^{2}$

$\sf \implies KE = \frac{1}{ \cancel{2}} \times \cancel{2} \times 3.4 \times 25$

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$\sf \implies KE = 85 \: J$

8 0

3 years ago

Read 2 more answers

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Dahasolnce [82]

Answer:

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Explanation:

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2 years ago

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A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the

Alex17521 [72]

Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3 m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1 V = 1.50*10^-3 m^3 ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

$dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5$

So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

$W = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J$

W = 3.12 J

Hope this helps!

4 0

3 years ago