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Doss [256]
3 years ago
12

A body accelerates by 25m/s 2 when it applied by 40n forces.what would be acceleration if it is applied by 80n forces

Physics
2 answers:
frutty [35]3 years ago
7 0

Answer:

50m/s2

Explanation:

since the force is doubled the acceleration will be doubled

il63 [147K]3 years ago
6 0

Answer:

50m/s²

Explanation:

cross muliplyy write down the values acceleration and force

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A 2.00-kg object A is connected with a massless string across a massless, frictionless pulley to a 3.00-kg object B. Object A re
slamgirl [31]

Answer:

  • tension: 19.3 N
  • acceleration: 3.36 m/s^2

Explanation:

<u>Given</u>

  mass A = 2.0 kg

  mass B = 3.0 kg

  θ = 40°

<u>Find</u>

  The tension in the string

  The acceleration of the masses

<u>Solution</u>

Mass A is being pulled down the inclined plane by a force due to gravity of ...

  F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N

Mass B is being pulled downward by gravity with a force of ...

  F = mg = (3 kg)(9.8 m/s^2) = 29.4 N

The tension in the string, T, is such that the net force on each mass results in the same acceleration:

  F/m = a = F/m

  (T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)

  T = (2(29.4) +3(12.5986))/5 = 19.3192 N

__

Then the acceleration of B is ...

  a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2

The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.

3 0
3 years ago
Consider a block on frictionless ice. Starting from rest, the block travels a distance din
sweet [91]

Answer:

<em>The distance is now 4d</em>

Explanation:

<u>Mechanical Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object.

The acceleration can be calculated by solving for a:

\displaystyle a=\frac{F}{m}

Once we know the acceleration, we can calculate the distance traveled by the block as follows:

\displaystyle d = vo.t+\frac{at^2}{2}

If the block starts from rest, vo=0:

\displaystyle d = \frac{at^2}{2}

Substituting the value of the acceleration:

\displaystyle d = \frac{\frac{F}{m}t^2}{2}

Simplifying:

\displaystyle d = \frac{Ft^2}{2m}

When a force F'=4F is applied and assuming the mass is the same, the new acceleration is:

\displaystyle a'=\frac{4F}{m}

And the distance is now:

\displaystyle d' = \frac{4Ft^2}{2m}

Dividing d'/d:

\displaystyle \frac{d' }{d}=\frac{\frac{4Ft^2}{2m}}{\frac{Ft^2}{2m}}

Simplifying:

\displaystyle \frac{d' }{d}=4

Thus:

d' = 4d

The distance is now 4d

3 0
3 years ago
If a bullet travels at 593.0 m/s, what is its speed in miles per hour?
Ksenya-84 [330]

We have 1 \; mile = 1609.34\; meters. So, 1 \; meter = \frac{1}{1609.34} \;mile.

1 \; hour = 3600 \; s. So 1 \; s = \frac{1}{3600}  \; hour.

Thus we can convert the units of the given quantity.

That is,

593\;m/s=593\;\frac{1/1609.34}{1/3600} \;miles/hour\\&#10;593\;m/s=1,326.51\;miles/hour.

The quantity is converted to the required units.


7 0
3 years ago
An athlete jumping vertically on a trampoline leaves the surface with a velocity of 8.5 m/s upward. what maximum height does she
Mumz [18]
<span>Her center of mass will rise 3.7 meters. First, let's calculate how long it takes to reach the peak. Just divide by the local gravitational acceleration, so 8.5 m / 9.8 m/s^2 = 0.867346939 s And the distance a object under constant acceleration travels is d = 0.5 A T^2 Substituting known values, gives d = 0.5 9.8 m/s^2 (0.867346939 s)^2 d = 4.9 m/s^2 * 0.752290712 s^2 d = 3.68622449 m Rounded to 2 significant figures gives 3.7 meters. Note, that 3.7 meters is how much higher her center of mass will rise after leaving the trampoline. It does not specify how far above the trampoline the lowest part of her body will reach. For instance, she could be in an upright position upon leaving the trampoline with her feet about 1 meter below her center of mass. And during the accent, she could tuck, roll, or otherwise change her orientation so she's horizontal at her peak altitude and the lowest part of her body being a decimeter or so below her center of mass. So it would look like she jumped almost a meter higher than 3.7 meters.</span>
8 0
3 years ago
1.)A tank travels at a rate of 10.0 km/hr for 12.00 minutes, then at 15.0 km/hr for 8.00
rosijanka [135]

12.00 min = 0.2 hr

8.00 min = 0.15 hr

Total distance:

(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) + (20.0 km/hr) (0.2 hr)

= 8.25 km

Average speed:

(10.0 km/hr + 15.0 km/hr + 20.0 km/hr) / 3

= 15 km/hr

Change in position:

(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) - (20.0 km/hr) (0.2 hr)

= 0.25 km

Average velocity:

(10.0 km/hr + 15.0 km/hr - 20.0 km/hr) / 3

≈ 1.67 m/s

8 0
3 years ago
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