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3 years ago

Which requires more work, lifting a 10-kg load a vertical distance of 2 m or lifting a 5-kg load a vertical distance of 4 m?

Physics

1 answer:

fredd [130]3 years ago

3 0

Both have the same work done.

5x4 = 20
10x2=20

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To convert minutes per second into kilometre per hour we multiply the speed with

ankoles [38]

Answer:

To convert m/sec into km/hr, multiply the number by 18 and then divide it by 5.

Explanation:

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3 years ago

A rope exerts a 280 N force while pulling an 80 Kg skier upward along a hill inclined at 12o. The rope pulls parallel to the hil

user100 [1]

Answer:

The speed of the skier after moving 100 m up the slope are of V= 25.23 m/s.

Explanation:

F= 280 N

m= 80 kg

α= 12º

μ= 0.15

d= 100m

g= 9,8 m/s²

N= m*g*sin(α)

N= 163 Newtons

Fr= μ * N

Fr= 24.45 Newtons

∑F= m*a

a= (280N - 24.5N) / 80kg

a= 3.19 m/s²

d= a * t² / 2

t=√(2*d/a)

t= 7.91 sec

V= a* t

V= 3.19 m/s² * 7.91 s

V= 25.23 m/s

4 0

3 years ago

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Kobotan [32]

Answer:

Explanation:

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2 years ago

A graph shows how the temperature of a substance changes as energy is added steadily over time.

OlgaM077 [116]

Answer:

A flat, horizontal line

Explanation:

A flat, horizontal line indicates a phase change.

The temperature does not increase because the added heat goes into converting one phase into another.

A is wrong. A downward-sloping line indicates that the temperature is decreasing with time.

C is wrong. An upward-sloping line indicates that the temperature is increasing with time.

8 0

3 years ago

Read 2 more answers

Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude

labwork [276]

Answer: $\tau=1.03\times 10^{-4}\ N-m$

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

$\tau=NIAB\ \sin\theta$

Let us assume that, $A=0.0008\ m^2$

$\theta$ is the angle between normal and the magnetic field, $\theta=90^{\circ}-30^{\circ}=60^{\circ}$

Torque is given by :

$\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m$

So, the net torque about the vertical axis is $1.03\times 10^{-4}\ N-m$ . Hence, this is the required solution.

4 0

3 years ago