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kirza4 [7]
3 years ago
8

ASAP PLEASE HELPPP

Physics
1 answer:
34kurt3 years ago
4 0

Answer: 3000k and centauri A

Explanation:

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In the demolition of an old building, a 1,300 kg wrecking ball hits the building at 1.07 m/s2. Calculate the amount of force at
Y_Kistochka [10]

Answer: F = 1391 N

Explanation:

The information given to you are:

Mass M = 1300 kg

Acceleration a = 1.07 m/s^2

The magnitude of the force striking the building will be

F = ma

Where

F = force

Substitute mass M and acceleration a into the formula

F = 1300 × 1.07

F = 1391 N

Therefore, the wrecking ball strikes the building with a force of 1391 N

3 0
3 years ago
An object is lifted from the surface of aspherical planet to an altitude equal to the radius of the planet.As a result, what hap
serg [7]

An object is lifted from the surface of a spherical planet to an altitude equal to the radius of the planet.  

As a result, the object's <em>mass remains the same</em>, and its <em>weight decreases</em> to 1/4 of whatever it is when the object is on the planet's surface.

8 0
3 years ago
A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of thebuilding. Ignore air resis
Bumek [7]

Answer:

a)3.5s

b)28.57m/S

c)34.33m/S

d)44.66m/S

Explanation:

Hello!

we will solve this exercise numeral by numeral

a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration

Y= VoT+0.5gt^{2}

where

Vo = Initial speed =0

T = time

g=gravity=9.81m/s^2

y = height=60m

solving for time

Y=0.5gt^2\\t=\sqrt{\frac{Y}{0.5g} } \\t=\frac{60}{0.5(9.81)} \\

T=3.5s

b)The horizontal speed remains constant since there is no horizontal acceleration. with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed

V=\frac{x}{t} =\frac{100}{3.5}=28.57m/s

c)

to find the final vertical velocity we use the equations for motion with constant velocity as follows

Vf=Vo+g.t    

Vf=0+(9.81 )(3.5)=34.335m/S          

d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares

V=\sqrt{Vx^2+Vy^2} =\sqrt{34.33^2+28.57^2} =44.67m/S

7 0
2 years ago
Describe how pieces of rope move as waves pass
MaRussiya [10]
Well, as the waves move it moves the rope as if its trying to take shape of it. Since the rope it light it will move along the ocean and the ocean will keep pushing up on the rope. (even without the waves the water is pushing the rope up so it can take its shape)

Maybe that'll help
6 0
3 years ago
A uniform stationary ladder of length L and mass M leans against a smooth vertical wall, while its bottom legs rest on a rough h
ikadub [295]

Answer:A uniform ladder of mass and length leans at an angle against a frictionless wall .If the coefficient of static friction between the ladder and the ground is , determine a formula for the minimum angle at which the ladder will not slip.

Explanation:A uniform ladder of mass and length leans at an angle against a frictionless wall .If the coefficient of static friction between the ladder and the ground is , determine a formula for the minimum angle at which the ladder will not slip.

3 0
3 years ago
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