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DochEvi [55]
3 years ago
12

Russell drags his suitcase 15.0 M from the door of his house to the car at a constant speed with a horizontal force of 95.0 N. H

ow much work does Russell do to overcome the force of
Physics
1 answer:
igomit [66]3 years ago
5 0

The work done is 1425 J

Explanation:

The work done by a force to move an object is given by

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and of the displacement

For the suitcase in this problem, we have:

F = 95.0 N is the force applied

d = 15.0 m is the displacement

\theta=0, since the force is parallel to the displacement

Substituting, we find

W=(95.0)(15.0)(cos 0)=1425 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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What were the two classifications of motion according to Aristotle?
wel

Answer:

natural motion and violent motion

Explanation:

8 0
2 years ago
using hooke's law, f spring = k triangle x, find the elastic constant of a spring that stretches 2 cm when a 4 newton force is a
Ksivusya [100]

As we know that spring force is given as

F = kx

here we know that

F = 4 N

x = 2 cm = 0.02 m

now from the above equation we will have

4 = k(0.02)

k = 200 N/m

so the elastic constant of the spring will be 200 N/m

8 0
3 years ago
In the diagram, R1 = 40.0 ,
Nostrana [21]

Answer:

51 Ω.

Explanation:

We'll begin by calculating the equivalent resistance of R₁ and R₃. This can be obtained as follow:

Resistor 1 (R₁) = 40 Ω

Resistor 3 (R₃) = 70.8 Ω

Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) =?

Since the two resistors are in parallel connection, their equivalent can be obtained as follow:

R₁ₙ₃ = R₁ × R₃ / R₁ + R₃

R₁ₙ₃ = 40 × 70.8 / 40 + 70.8

R₁ₙ₃ = 2832 / 110.8

R₁ₙ₃ = 25.6 Ω

Finally, we shall determine the equivalent resistance of the group. This can be obtained as follow:

Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) = 25.6 Ω

Resistor 2 (R₂) = 25.4 Ω

Equivalent Resistance (Rₑq) =?

Rₑq = R₁ₙ₃ + R₂ (series connection)

Rₑq = 25.6 + 25.4

Rₑq = 51 Ω

Therefore, the equivalent resistance of the group is 51 Ω.

4 0
2 years ago
. A bathysphere used for deep-sea exploration has a radius of 1.50 m and a mass of 1.20 104 kg. To dive, this submarine takes on
dangina [55]

Answer:

Explanation:

Given that,

Bathysphere radius

r = 1.5m

Mass of bathysphere

M = 1.2 × 10⁴ kg

Constant speed of descending.

v = 1.2m/s

Resistive force

Fr = 1100N upward direction

Density of water

ρ = 1.03 × 10³kg/m³

The volume of the bathysphere can be calculated using

V = 4πr³ / 3

V = 4π × 1.5³ / 3

V = 14.14 m³

The Bouyant force can be calculated using

Fb = ρgV

Fb = 1.03 × 10³ × 9.81 × 14.14

Fb = 142,846.18 N

Buoyant force is acting upward

Weight of the bathysphere

W = mg

W = 1.2 × 10⁴ × 9.81

W = 117,720 N

Weight is acting downward

The net positive buoyant using resolving

Fb+ = Fb — W

Fb+ = 142,846.18 — 117,720

Fb+ = 25,126.18 N

The force acting downward is the weight of the submarine and it is equal to the positive buoyant force and the resistive force

W = Fb+ + Fr

W = 25,126.18 + 1100

W = 26,226.18

mg = 26,226.18

m = 26,226.18 / 9.81

m = 2673.4kg

Mass of submarine is 2673.4kg

8 0
2 years ago
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