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expeople1 [14]
3 years ago
5

Which are consider to be simple machines? Select all that apply.

Physics
1 answer:
Sedbober [7]3 years ago
7 0

Answer:

b

Explanation:

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Different masses are hung on a spring scale calibrated in newtons. The force exerted by gravity on 1.0 kg is shown in the image
nekit [7.7K]

d.88 n

Explanation:

The force exerted by gravity on the mass is 88N;

Gravitational force is usually experienced by two bodies with masses.

This force causes all objects on earth to accelerate at 9.8m/s².

    Force of gravity = mass x acceleration due to gravity

In this case, the force exerted on the object by gravity will be the weight of the object.

Weight is a force applied gravity on a body:

   Weight of the masses = Force of gravity = mass x acceleration

   Given mass = 9kg

             Force of gravity = 9 x 9.8 = 88.2N approximately 88N

learn more:

Weight and mass brainly.com/question/5956881

#learnwithBrainly

5 0
3 years ago
A parachute falling to the ground.<br><br>​
butalik [34]

Answer:

a parachute falling to the ground is uniform

7 0
2 years ago
A boat sails south with the help of a wind blowing in the direction S36°E with magnitude 300 lb. Find the work done by the wind
goldfiish [28.3K]

Answer:

The work done by the wind as the boat moves 130 ft is (rounded) W= 31,550 ft-lb.

Explanation:

F= 300 lb < -54º

Fsouth= 300 lb * cos(36º)

Fsouth= 242.7 lb

d= 130 ft

W= F*d

W= 31551 ft-lb

6 0
3 years ago
Write a hypothesis for Part II of the lab, which is about the relationship described by F = ma. In the lab, you will use a toy c
Temka [501]

Answer:

F=ma is the relationship where, F is force, m is mass and a is acceleration.

Newton's second law states that  the unbalanced force applied to the object accelerates the object which is directly proportional to the force and inversely to the mass.

If we apply force to a toy car then It will accelerate.

This is how Newton's second law of motion is verified.

6 0
3 years ago
Read 2 more answers
According to the Guinness Book of World Records, the longest home run ever measured was hit by Roy "Dizzy" Carlyle in a minor-le
Alina [70]
Range of a projectile motion is given by
R = v cos θ / g (v sin θ + sqrt(v^2 sin^2 θ + 2gy_0)); where R = 188m, θ = 41°, g = 9.8m/s^2, y_0 = 0.9

188 = v cos 41° / 9.8 (v sin 41° + sqrt(v^2 sin^2 41° + 2 x 9.8 x 0.9)) = 0.07701(0.6561v + sqrt(0.4304 v^2 + 17.64)) = 0.05053v + 0.07701sqrt(0.4304v^2 + 17.64)
0.07701sqrt(0.4304v^2 + 17.64) = 188 - 0.05053v
0.005931(0.4304v^2 + 17.64) = 35344 - 19v + 0.002553v^2
0.002553v^2 + 0.1046 = 35344 - 19v + 0.002553v^2
19v = 35344 - 0.1046 = 35343.8954
v = 35343.8954/19 = 1860 m/s
5 0
3 years ago
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