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3 years ago

Book sliding across a frictionless table would never stop moving, even though there is

Physics

1 answer:

Lina20 [59]3 years ago

4 0

Answer:

Yes, this is true, unless there were other outside forces, such as wind.

Explanation:

The First Law of Motion states, "A body at rest will remain at rest, and a body in motion will remain in motion unless it is acted upon by an external force."

Hope this helped!! :)

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A student measures the time it takes for two reactions to be completed reaction eight is completed and 57 seconds and reaction b

Maslowich

The answer to this problem is 1200

8 0

3 years ago

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

8 0

3 years ago

Is it possible for two different atoms of carbon to have different numbers of neutrons in their nuclei? Explain.

bezimeni [28]

Answer:

Yes , it is possible for two different atoms of carbon to have different numbers of neutrons in their nuclei .

Explanation:

Isotopes -

Atoms of the element with same number of the electrons and protons , but differ in the number of neutrons , are called as the isotopes .

The isotopes of the element have the same number of protons and electrons , hence have the same chemical and physical properties .

Many isotopes occurs naturally .

In case of Carbon ,

Carbon too have isotopes ,

i.e.

Carbon - 13 and Carbon - 14

carbon - 13 , have seven neutrons

Carbon - 14 , have eight neutrons .

4 0

3 years ago

An electric vehicle starts from rest and accelerates at a rate of 2.4 m/s2 in a straight line until it reaches a speed of 27 m/s

DENIUS [597]

A) use v=u+at for both

First section, v=27, u=0, a=2.4. You should get 11seconds.
Second section, v=0, u=27, a=-1.3. You should get 21seconds.
This means that the total time is 22seconds.

b) You can either use s=ut+0.5at^2 or v^2=u^2+2as. Personally, I would use the second one as you are not relying on your previous answer.

First section, v=27, u=0, a=2.4. You should get 152m.
Second section, v=0, u=27, a=-1.3. You should get 280m.
This makes your overall displacement 432m.

6 0

3 years ago

There is a 250-m-high cliff at half dome in yosemite national park in california. suppose a boulder breaks loose from the top of

lorasvet [3.4K]

Part A. For this part, we use two equations for linear motion:

y = y0 + v0 t + 0.5 g t^2 ---> 1

vf = v0 + g t ---> 2

First we solve for t using equation 1: y0 = 0 (initial point at top), y = 250 m, v0 = 0 (at rest)

250 = 0.5 (9.8) t^2

t = 7.143 s

Now we solve for final velocity vf using equation 2:

vf = g t

vf = 9.8 (7.143)

vf = 70 m/s

Part B. First we solve for the time it takes for the sound to reach the tourist.

t(sound) = 250 / 335 = 0.746 s

Therefore the total time would be:

t = 0.746 s + 0.300 s

t = 1.05 s

Hence there is enough time for the tourist to get out before the boulder hits him.

7 0

3 years ago

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