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2 years ago

A box slides down a frictionless incline, gaining speed. The work done by the normal force n is _______.

2 answers:

7 0

The work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

<h3>What is normal force?</h3>

The force of contact is called the normal force. When the two surfaces are in contact with each other, then the normal force acts.

This force is applied by the solid bodies on each other in order to prevent the passing through each other.

A box slides down a frictionless incline, gaining speed. For this box, the value of work done by normal force has to be found out. Let's analyze the given condition.

The body is gaining the speed, which means there is a change in kinetic energy.
The change in kinetic energy is equal to the work done.
The friction force is the product of coefficient of the friction and normal force.
The friction force for the given case is zero. Thus, the normal force must be equal to the zero.

Thus, the work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

Learn more about the normal force here;

brainly.com/question/10941832

Zarrin [17]2 years ago

4 0

Answer:

Explanation:

The normal force will be perpendicular to the motion. A zero-work situation is one where "a force everywhere perpendicular to the motion does no work" so according to this principle, no work is done.

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During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 6.82 s, how high does it rise? The accel

scoray [572]

The answer is 56.98 m

The dislocation (d) is:
d = v1 * t + 1/2 * a * t²
v1 - initial velocity
t - time
a - acceleration of gravity

We know:
d = ?
v1 = ?
t = 6.82 s / 2 = 3.41 (it reaches the peak at half time)
a = - 9.8 m/s²

Let's first calculate v1:
v2 = v1 + at
v2 - final velocity (v2 = 0 when it reaches peak)

0 = v1 + -9.8 * 3.41
0 = v1 - 33.418

v1 = 33.418 m/s

d = v1 * t + 1/2 * a * t²
d = 33.418 * 3.41 + 1/2 * -9.8 * 3.41²
d = 113.96 - 56.98
d = 56.98 m

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3 years ago

Explain how heat is related to temperature and thermal energy...

daser333 [38]

Answer:

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3 years ago

If the jet is moving at a speed of 1140 km/h at the lowest point of the loop, determine the minimum radius of the circle so that

AnnZ [28]

To solve this problem we will apply the concepts related to the centripetal Force and the Force given by weight and formulated in Newton's second law. Through the two expressions we can find the radius of curve made in the hand. To calculate the normal force, we will include the concepts of sum of forces to obtain the net force on the body at the top and bottom of the maneuver. The expression for centripetal force acting on the jet is

$F_c = \frac{mv^2}{r}$

According to Newton's second law, the net force acting on the jet is

F = ma

Here,

m = mass

a = acceleration

v = Velocity

r = Radius

PART A ) Equating the above two expression the equation for radius is

$\frac{mv^2}{r} = ma$

$r = \frac{v^2}{a}$

Replacing with our values we have that

$r = \frac{(1140km/hr[\frac{1000m}{1km}\frac{1hour}{3600s}])^2}{7(9.8m/s^2)}$

$r = 1.462*10^3m$

PART B )

- The expression for effective weight of the pilot at the bottom of the circle is

$N = mg +\frac{mv^2}{r}$

$N = (69kg)(9.8m/s^2)+\frac{(69)(1140km/hr[\frac{1000m}{1km}\frac{1hour}{3600s}])^2}{1.462*10^3m}$

$N = 5408.87N$

Note that the normal reaction N is directed upwards and gravitational force mg is directed downwards. At the bottom of the circle, the centripetal force is directed upwards. So the centripetal force is obtained from the gravitational force and the normal reaction.

- The expression for effective weight of the pilot at the top of the circle is

$N = mg -\frac{mv^2}{r}$

$N = (69kg)(9.8m/s^2)-\frac{(69)(1140km/hr[\frac{1000m}{1km}\frac{1hour}{3600s}])^2}{1.462*10^3m}$

$N = 4056.47N$

Note that at the top of the circle the centripetal force is directed downwards. So the centripetal force is obtained from normal reaction and the gravitational force.

4 0

3 years ago

Use Kepler’s third law and the orbital motion of Earth to determine the mass of the Sun. The average distance between Earth and

Anna11 [10]

Kepler’s third law formula: T^2=4pi^2*r^3/(GM)

We’re trying to find M, so:

M=4pi^2*r^3/(G*T^2)

M=4pi^2*(1.496 × 10^11 m)^3/((6.674× 10^-11N*m^2/kg^2)*(365.26days)^2)

M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))

Let’s work with the units:

(m^3)/((N*m^2/kg^2)*days^2))=

=(m^3*kg^2)/(N*m^2*days^2)

=(m*kg^2)/(N*days^2)

=(m*kg^2)/((kg*m/s^2)*days^2)

=(kg)/(days^2/s^2)

=(kg*s^2)/(days^2)

So:

M=1.48× 10^40(kg*s^2)/(days^2)

Now we need to convert days to seconds in order to cancel them:

1 day=24 hours=24*60minutes=24*60*60s=86400s

M=1.48× 10^40(kg*s^2)/((86400s)^2)

M=1.48× 10^40(kg*s^2)/( 86400^2*s^2)

M=1.48× 10^40kg/86400^2

M=1.98x10^30kg

The closest answer is 1.99 × 10^30

(it may vary a little with rounding – the difference is less than 1%)

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3 years ago

Read 2 more answers

PLEASE HELP THIS IS TIMED!

Monica [59]

3.5 because yea periodt

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3 years ago