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Allisa [31]
3 years ago
15

What is the formula of moment of force​

Physics
1 answer:
Nat2105 [25]3 years ago
4 0
Moment of force=fxd

Explain
M=fxd
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a body dropped from a height reaches a velocity of 13m/s just before touching the ground. What is the initial height of the ball
SashulF [63]

Hi there!

We can use the following (derived) equation to solve for the final velocity given height:

vf = √2gh

We can rearrange to solve for height:

vf² = 2gh

vf²/2g = h

Plug in the given values (g = 9.81 m/s²)

(13)²/2(9.81) = 8.614 m

We can calculate time using the equation:

vf = vi + at, where:

vi = initial velocity (since dropped from rest, = 0 m/s)

a = acceleration (in this instance, due to gravity)

Plug in values:

13 = at

13/a = t

13/9.81 = 1.325 sec

5 0
2 years ago
A 7.00 kg bowling ball is held 2.00 m above the ground. Using g = 9.80 m/s2, how much energy does the bowling ball have due to i
deff fn [24]
mgh =7*2*9.8=137.2. It should be the answer.
4 0
3 years ago
Read 2 more answers
Ls -2 a solution of 4x +3= -5?.<br>​
torisob [31]

Answer:

yes

Explanation:

Let's solve your equation step-by-step.

4x+3=−5

Step 1: Subtract 3 from both sides.

4x+3−3=−5−3

4x=−8

Step 2: Divide both sides by 4.

4x  / 4  =  −8  / 4

x=−2

Hope it helps,

Please mark me as the brainliest

Thank you

5 0
3 years ago
A ball is thrown vertically upwards from the edge of the cliff and hits the ground at the base of the cliff with a speed of 30 m
olya-2409 [2.1K]

To solve this problem we will apply the linear motion kinematic equations. From the definition of the final velocity, as the sum between the initial velocity and the product between the acceleration (gravity) by time, we will find the final velocity. From the second law of kinematics, we will find the vertical position traveled.

v = v_0 -gt

Here,

v = Final velocity

v_0 = Initial velocity

g = Acceleration due to gravity

t = Time

At t = 4s, v = -30m/s (Downward)

Therefore the initial velocity will be

-30 = v_0 -9.8(4)

v_0 = 9.2m/s

Now the position can be calculated as,

y = h +v_0t -\frac{1}{2}gt^2

When it has the ground, y=0 and the time is t=4s,

0 = h+(9.2)(4)-\frac{1}{2} (9.8)(4)^2

h = 41.6m

Therefore the cliff was initially to 41.6m from the ground

7 0
2 years ago
A golf club hits a 0.04551 kg golf ball off a golf tee. The club is in contact with the ball for 0.020 s, and the force applied
Juliette [100K]

Answer:

v = 50.5 m/s

Explanation:

F = (m)(^v/^t)

115N = (0.04551kg)(v/(0.020s))

2,526.917161 m/s² = v/(0.020s)

v = 50.53834322 m/s

v = 50.5 m/s

8 0
3 years ago
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