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3 years ago

A 1,300 kg wrecking ball hits the building at 1.07 m/s2.

Physics

2 answers:

dezoksy [38]3 years ago

7 0

So we know F=ma
m means mass and a means acceleration
so Force= ma
so F=1300X1.07=1391N
hope you like it
if it helps you please mark the answer as brainliest answer

tatuchka [14]3 years ago

6 0

F (force) = mass × acceleration.

The unit N (Newton) is known as kilogram-meter/seconds2.

Thus, F = 1300 kg × 1.07 m/s2 = 1391 N.

The answer is 1391 N.

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A 215-kg load is hung on a wire of length of 3.60 m, cross-sectional area 2.00 10-5 m2, and Young's modulus 8.00 1010 N/m2. What

Y_Kistochka [10]

Answer:

4.74 * 10^-3

Explanation:

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3 years ago

A projectile is launched horizontally from a height of 8.0 m. The projectile travels 6.5 m before hitting the ground.

spayn [35]

Answer:

5.09 m/s

Explanation:

Use the height to find the time it takes to land:

y = y₀ + v₀ᵧ t + ½ gt²

0 = 8.0 m + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 1.28 s

Now use the horizontal distance to find the initial velocity.

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v₀ = 5.09 m/s

7 0

3 years ago

9. The three types of stress that act on Earth's rocks are compression, tension, and A. strain. B. shear. C. tephra. D. shale.

Evgen [1.6K]

The three main types of stress in a rock are shearing, tension, and compression. hope. that helped

5 0

3 years ago

Read 2 more answers

What is the strength of the electric field inside the membrane just before the action potential?

gtnhenbr [62]

Answer:

Incomplete question, check attachment for the graph needed to solve problem.

A 8.1nm........

Explanation:

Electric Field is given as

E=V/d

Where V is voltage

And d is the distance apart

E is the electric field

The voltage V just before action of potential is -70mV,

The value d=8.1nm

d=8.1×10^-9m

E=V/d

E=-70×10^-3/8.1×10^-9

E=-8.6×10^6 N/C

Then the magnitude of the electric field is 8.6×10^6N/C

5 0

3 years ago

You are a member of an alpine rescue team and must project a box of supplies, with mass m, up an incline of constant slope angle

AlekseyPX

Answer:

$v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}$

Explanation:

When we push the box from the bottom of the incline towards the top then by work energy theorem we can say that

Work done by all the forces = change in kinetic energy of the system

$- mgh - F_f (s) = 0 - \frac{1}{2}mv^2$

here we know that

$F_f = \mu_k mg cos\theta$

also we know that the length of the incline is given as

$s = \frac{h}{sin\theta}$

now we have

$- mgh - \mu_k mgcos\theta(\frac{h}{sin\theta}) = -\frac{1}{2}mv^2$

so we have

$v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}$

3 0

3 years ago