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3 years ago

If a force of 50 Newton’s was applied to an object with a mass of 500 grams, what will the objects acceleration be?

Physics

1 answer:

lapo4ka [179]3 years ago

5 0

Answer: 100 m/s^2

F=ma

Explanation:

50N = 50 kg*m/s^2

500g = 0.5 kg

F=ma

a = F/m

a = (50 kg*m/s^2)/(0.5 kg)

a = 100 m/s^2

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A plane passes over Point A with a velocity of 8000 m/s north. Forty seconds later it passes over Point B at a velocity of 10,00

Airida [17]

Answer:

The planes’ acceleration from A to B is 500m/s^2

Explanation:

Given that the initial velocity u is 8000m/s

and also given the final velocity v=10,000 m/s

the time taken to move from A to B = 40 second

The acceleration is defined as the rate of change of velocity with time

we know that the expression for acceleration is given as

a=(v-u)/t

substituting our given data into the expression for a we have

a=(10000-8000)/40

a=2000/40

a=500m/s^2

The planes’ acceleration from A to B is 500m/s^2

7 0

3 years ago

A toy train is pushed forward and released at Xi = 4.0 m with a speed of 2.5 m/s. It rolls at a steady speed for 2.0 s, then one

Fittoniya [83]

I don't think that 4m has anything to do with the problem. anyway. here. A___________________B_______C where A is the point that the train was released. B is where the wheel started to stick C is where it stopped From A to B, v=2.5m/s, it takes 2s to go A to B so t=2 AB= v*t = 2.5 * 2 = 5m The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m then BC= AC-AB = 7.7-5 = 2.7m now consider BC v^2=u^2+2as where u is initial speed, in this case is 2.5m/s v is final speed, train stop at C so final speed=0, so v=0 a is acceleration s is displacement, which is BC=2.7m substitute all the number into equation, we have 0^2 = 2.5^2 + 2*a*2.7 0 = 6.25 + 5.4a a = -6.25/5.4 = -1.157 so acceleration is -1.157m/(s^2)

8 0

3 years ago

Read 2 more answers

A car starts from rest and accelerates uniformly at 2.0 m/s2 toward the north. A second car starts from rest 4.0 s later at the

yKpoI14uk [10]

Answer:

the correct solution is 13 s

Explanation:

This is a kinematic problem, let's use accelerated rectilinear motion relationships.

For the first car it has an accelerometer of 2.0 m/s²

x = v₀₁ t + ½ a₁ t²

The second car leaves the same point, but 4.0 seconds later

x = v₀₂ (t-4) + ½ a₂ (t-4)²

With this form we use the same time for both cars.

The initial speeds are zero for both vehicles leave the rest, at the point where they are located has the same position

x = ½ a₁ t²

x = ½ a₂ (t-4)²

Let's solve

a₁ t² = a₂ (t-4)²

a₁/a₂ t² = t² -2 4 t + 16

t² (1- 2.0 / 4.0) - 8 t +16

t² 0.5 - 8 t +16 = 0

t² -16 t + 32 = 0

Let's solve the second degree equation

t = [16 ±√( 16² - 4 32)] / 2

t = ½ (16 ± 11,3)

Solutions

t1 = 13.66 s

t2 = 2.34 s

These are the mathematical solutions for the meeting point, but car 2 leaves after 4 seconds, so the only solution is 13.66 s

the correct solution is 13 s, if you have to select one the nearest 12s

6 0

3 years ago

A student is trying to determine the acceleration of a feather as she drops it to the ground. if the student is looking to achie

Anna [14]

The coordinate system should have the origin at the point where the feather is dropped and the downward direction is to be taken as positive.

All falling bodies experience acceleration towards the center of the Earth due to the force of gravitational attraction exerted on the object by the Earth. A feather, when dropped experiences an acceleration in the downward direction. Since the acceleration of the feather is in the downward direction, a feather, when dropped with zero initial velocity, has its velocity vector directed in the direction of its acceleration.

If the downward direction is taken as positive, the falling feather can be said to have a positive velocity and a positive acceleration.

5 0

3 years ago

Options: A.) 10 N B.) 15 N C.) 25 N D.) 35N

Lady_Fox [76]

Given:

F_gravity = 10 N

F_tension = 25 N

Let's find the net centripetal force exterted on the ball.

Apply the formula:

$\sum ^{}_{}F_{\text{net}}=F_1+F_2=F_{centripetal}$

From the given figure, the force acting towards the circular path will be positive, while the force which points directly away from the center is negative.

Hence, the tensional force is positive while the gravitational force is negative.

Thus, we have:

$F_{\text{net}}=F_{\text{centripetal}}=F_{tension}-F_{gravity}=25N-10N=15N$

Therefore, the net centripetal force exterted on the ball is 15 N.

ANSWER:

15 N

7 0

1 year ago