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3 years ago

A 25kg box fell 200m with an acceleration of 5 m/s2. with what force did it hit the floor when it landed?

1 answer:

6 0

According to Newton's Second Law of motion, the net force acting on the object is equal to its mass multiplied by its acceleration. In formula, it is written as

Net Force =mass * acceleration
Net force = 25 kg * 5m/s^2
Net force = 125 Newtons

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You lift a 10-kg box to a height of 1m. How much work do you do on the box when you lift it from the ground? (g= 9.81 m/s2)

Andreyy89

Answer:

98.1 Joule

Explanation:

Solution,

⇒Mass(m)=10kg

⇒Weight(F)=Mg=10×9.81=98.1N

⇒Distance(d)=1m

Now,

Work done=F×d

=98.1×1

=98.1J

5 0

2 years ago

A 1kg box is pushed on a flat surface that is 250m long. The box is initially at rest and then pushed with a constant Net force

sasho [114]

Answer:

C) 50 m/s

Explanation:

With the given information we can calculate the acceleration using the force and mass of the box.

Newton's 2nd Law: F = ma

5 N = 1 kg * a
a = 5 m/s²

List out known variables:

v₀ = 0 m/s
a = 5 m/s²
v = ?
Δx = 250 m

Looking at the constant acceleration kinematic equations, we see that this one contains all four variables:

v² = v₀² + 2aΔx

Substitute known values into the equation and solve for v.

v² = (0)² + 2(5)(250)
v² = 2500
v = 50 m/s

The final velocity of the box is C) 50 m/s.

7 0

3 years ago

You create a ramp using two text books and a 0.50m board. Using a timer you determine that a cart can roll down the ramp in 0.55

ahrayia [7]

Answer:

The velocity of the cart at the bottom of the ramp is 1.81m/s, and the acceleration would be 3.30m/s^2.

Explanation:

Assuming the initial velocity to be zero, we can obtain the velocity at the bottom of the ramp using the kinematics equations:

$v=v_0+at\\\\v^2=v_0^2+2ad$

Dividing the second equation by the first one, we obtain:

$v=\frac{v_0^2+2ad}{v_0+at}$

And, since $v_0=0$ , then:

$v=\frac{2ad}{at}\\\\v=\frac{2d}{t}\\\\v=\frac{2(0.50m)}{0.55s}\\\\v=1.81m/s$

It means that the velocity at the bottom of the ramp is 1.81m/s.

We could use this data, plus any of the two initial equations, to determine the acceleration:

$v=v_0+at\\\\\implies a=\frac{v}{t}\\\\a=\frac{1.81m/s}{0.55s}\\\\a=3.30m/s^2$

So the acceleration is 3.30m/s^2.

7 0

3 years ago

A car traveling at 60km/h undergoes uniform acceleration at a rate of 2/ms^2 until is velocity reached 120km/h determine the dis

jenyasd209 [6]

Explanation:

Given that,

Initial speed of a car, u = 60 km/h = 16.67 m/s

Acceleration, a = 2m/s²

Final speed, v = 120 km/h = 33.33 m/s

We need to find the distance traveled and the time taken to make the distance.

acceleration = rate of change of velocity

$a=\dfrac{v-u}{t}\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{33.33 -16.67 }{2}\\\\t=8.33\ s$

let the distance be d.

$d=\dfrac{v^2-u^2}{2a}\\\\d=\frac{33.33^{2}-16.67^{2}}{2(2)}\\\\d=208.25\ m$

Hence, the distance traveled and the time taken to make the distance is 208.25 m and 8.33 seconds respectively.

3 0

2 years ago

How do you work out how many atoms there are in a isotope?

GalinKa [24]

206Pb = 1.342 x10^22 atoms

To find the number of atoms, you must first find the number of moles. If 238U is 238.029g/mol, and we have 1.75 grams, how many moles is that? 1.75 divided by 238.029 = 0.007352045 moles. To find the number of atoms in 0.007352045 moles, you multiply by a mole: 
0.007352045 x 6.02 x 10^23 = 4.426 x10^21 atoms. 

Same procedure for 206Pb: 
4.59 divided by 205.97446 = 0.022284316 moles 
0.022284316 x 6.02 x 10^23 = 1.342 x10^22 atoms. 

Hope that helps you!
https://answers.yahoo.com/question/index?qid=20100331153014AAoMXcu

4 0

3 years ago