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kvasek [131]
3 years ago
9

According to Newton’s first law of motion what will an object in motion do when no external force acts on it?

Physics
1 answer:
PIT_PIT [208]3 years ago
4 0
The object will continue moving in a straight line at constant speed.
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the body moving with uniform velocity has acceleration zero.

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3 years ago
A ski gondola is connected to the top of a hill by a steel cable of length 620 m and diameter 1.5 cm. As the gondola comes to th
xz_007 [3.2K]

Answer:

(a) 89 m/s

(b) 11000 N

Explanation:

Note that answers are given to 2 significant figures which is what we have in the values in the question.

(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.

v=\dfrac{2\times 620 \text{ m}}{14\text{ s}} = \dfrac{1240\text{ m}}{14\text{ s}}=88.571428\ldots \text{ m/s}= 89\text{ m/s}

(b) The tension, T, is given by

v =\sqrt{\dfrac{T}{\mu}}

where v is the speed, T is the tension and \mu is the mass per unit length.

Hence,

T = \mu\cdot v^{2}

To determine \mu, we need to know the mass of the cable. We use the density formula:

\rho = \dfrac{m}{V}

where m is the mass and V is the volume.

m=\rho\cdot V

If the length is denoted by l, then

\mu = \dfrac{m}{l} = \dfrac{\rho\cdot V}{l}

T = \dfrac{\rho\cdot V}{l} v^{2}

The density of steel = 8050 kg/m3

The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is

V = \pi \dfrac{d^{2}}{4} l

T = \dfrac{\rho\cdot\pi d^2 l}{4l}v^2 = \dfrac{\rho\cdot\pi d^2}{4}v^2

T = \dfrac{8050\times\pi\times0.015^2}{4} \times 88.57^2

T = 11159.4186\ldots \text{ N} = 11000 \text{ N}

4 0
3 years ago
A horizontal force of magnitude 30.2 N pushes a block of mass 3.50 kg across a floor where the coefficient of kinetic friction i
marshall27 [118]

Answer:

A) 89.39 J

B) 30.39J

C) 23.8 J

Explanation:

We are given;

F = 30.2N

m = 3.5 kg

μ_k = 0.646

d = 2.96m

ΔEth (Block) = 35.2J

A) Work done by the applied force on the block-floor system is given as;

W = F•d

Thus, W = 30.2 x 2.96 = 89.39 J

B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;

ΔEth = μ_k•mgd

Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J

Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.

Thus,

ΔEth = ΔEth (Block) + ΔEth (floor)

Thus,

ΔEth (floor) = ΔEth - ΔEth (Block)

ΔEth (floor) = 65.59J - 35.2J = 30.39J

C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;

W = K + ΔEth

Therefore;

K = W - ΔEth

K = 89.39 - 65.59 = 23.8J

3 0
3 years ago
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