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4 years ago

What are two components that make up force?

Physics

2 answers:

Sveta_85 [38]4 years ago

6 0

The answer is mass and acceleration

natka813 [3]4 years ago

5 0

Gravity and energy hope this helps bro

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Human centrifuges are used to train military pilots and astronauts in preparation for high-g maneuvers. A trained, fit person we

jeka57 [31]

(a) $4.14 rad/s^2$

The relationship beween centripetal acceleration and angular speed is

$a=\omega^2 r$

where

$\omega$ is the angular speed

r is the radius of the circular path

Here we gave

$a = 9g = 88.2 m/s^2$ is the centripetal acceleration

r = 5.15 m is the radius

Solving for $\omega$ , we find:

$\omega = \sqrt{\frac{a}{r}}=\sqrt{\frac{88.2 m/s^2}{5.15 m}}=4.14 rad/s^2$

(b) 21.3 m/s

The relationship between the linear speed and the angular speed is

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular speed

r is the radius of the circular path

In this problem we have

$\omega=4.14 rad/s$

r = 5.15 m

Solving the equation for v, we find

$v=(4.14 rad/s)(5.15 m)=21.3 m/s$

7 0

3 years ago

Read 2 more answers

Water pressurized to 3.5 x 105 Pa is flowing at 5.0 m/s in a horizontal pipe which contracts to 1/3 its former area. What are th

masha68 [24]

Answer:

the pressure after contraction is 2×10^5 Pa

the speed after contraction is 15m/s

Explanation:

We were given Pressure P to be 3.5 x 10^5 that is Flowing with speed of 5.0 m/s,

For us to calculate pressure we need to calculate the area first as;

Let initial Area = A₁

And Final area A₂

We were told that in a horizontal pipe it contracts to 1/3 its former area. Which means

A₂= A₁/3.................

V₁ is the speed

the pressure and speed of the water after the contraction can be calculated using equation of continuity below

A₂V₂ = A₁V₁

But

If we substitute given value in the expresion we have

V₂ = (3A *5)/A

V₂ = 15m/s

Therefore, the speed after contraction is 15m/s

Now we can calculate the pressure using

Bernoulli's equation

p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂

But we know that the pipe is horizontal, then "h" terms cancel out then

p₁ + ½ρv₁² = p₂ + ½ρv₂²

Making P₂ subject of formula we have

p₂ = 0.5ρ( V ₁² - v₂² ) + P₁

P₂=. 0.5 × 1000 (5² -15² ) + 3*10^5

=2×10^5 Pa

Therefore, the pressure after contraction is 2×10^5 Pa

7 0

3 years ago

Help ?

Katyanochek1 [597]

I think it would be the one talking about if there’s water there would still be energy because water is used as a source of energy because there’s so much of it and it can be used again and again

8 0

3 years ago

By the 2nd century BCE the ____ democratized astrology<br> A. Greeks<br> B. Egyptians<br> C. Romans

elena-14-01-66 [18.8K]

Answer:

Explanation:

3 0

3 years ago

PHYSICS PLEASE HELP 80 PTS!!!!!!

VladimirAG [237]

1. B) 32 m/s

The speed of the baseball is given by

$v=\frac{d}{t}$

where

d = 48 m is the distance travelled

t = 1.5 s is the time taken

Substituting,

$v=\frac{48}{1.5}=32 m/s$

2. C) 12 m/s

The average velocity is given by

$v=\frac{d}{t}$

where

d = 60 - 0 = 60 m is the displacement

t = 5 s is the time interval

Substituting,

$v=\frac{60}{5}=12 m/s$

3. Missing diagram

4. D) velocity

In fact, velocity is defined as the ratio between the displacement ( $\Delta s$ ) and the time interval ( $\Delta t$ ) needed to achieve that change in position:

$v=\frac{\Delta s}{\Delta t}$

Similarly, acceleration is defined as the ratio between the change in velocity ( $\Delta v$ ) and the time interval ( $\Delta t$ ):

$a=\frac{\Delta v}{\Delta t}$

Comparing the two equations, we see that displacement is to velocity as velocity is to acceleration.

5. C) 5/1 cm/year

We know that the ridge moves 25 cm in 5 years, so we have:

- Displacement (d): d = 25 cm

- Time interval (t): t = 5 y

Using the equation for the velocity:

$v=\frac{d}{t}=\frac{25 cm}{5 y}=5 cm/y$

6. D

The missing graph is attached. Each graph represents a distance (on the y-axis, measured in metres) versus a time (on the x-axis, measured in seconds). Therefore, the speed in each graph can be simply calculated from the slope of the curve, since speed is the ratio between distance and time:

$v=\frac{\Delta y}{\Delta x}$

For each graph:

A. $\frac{\Delta y}{\Delta x}=\frac{8.6-2}{10}\sim 0.66$

B. $\frac{\Delta y}{\Delta x}=\frac{5.7-0}{10}\sim 0.57$

C. $\frac{\Delta y}{\Delta x}=\frac{9.6-3}{10}\sim 0.66$

D. $\frac{\Delta y}{\Delta x}=\frac{7.5-0}{5}\sim 1.5$

So we can say that the correct graph must be graph D, the closest to 1.57.

7. B) The car has come to a stop and has zero velocity

The missing graph is attached.

The graph shows the position of the car versus time. From the graph, we can see that in segment B, the position of the car does not change: this means that its velocity is zero, since the velocity is defined as the ratio between the change in position (displacement) and the time interval:

$v=\frac{\Delta s}{\Delta t}$

And since the displacement is zero, the velocity is also zero.

8. C) from 4.5 to 5.0 seconds

Missing graph is attached.

The graph represents the distance covered versus the time taken: therefore, the speed of the ball can be evaluated from the slope of the curve.

The only region in which the slope of the curve is zero is between 4.5 seconds and 5.0 seconds: therefore, this is the region where the soccer ball is not moving.

9. D) diagonal line with varying slope, from 3 to 4.

The table of data is:

Time (s) 0 1 2 3 4

Distance (m) 0 3 6 12 16

We want to know how the distance/time graph would like like. We have:

- At t = 1, the distance is $d=3$ , so the slope is $\frac{d}{t}=\frac{3}{1}=3$

- Similarly, at t = 2: $\frac{d}{t}=\frac{6}{2}=3$

- Instead, at t =3, the slope is $\frac{d}{t}=\frac{12}{3}=4$

- Similarly, at t=4, $\frac{d}{t}=\frac{16}{4}=4$

So the correct answer is D.

10. D) To move in a circle requires an acceleration and therefore a net force. This force is supplied by the slingshot.

For an object in circular motion, the velocity constantly changes (because the direction changes): this means that an acceleration is required (towards the centre of the circle), and therefore a force must be exerted (also towards the centre of the circle) to keep the object in the circular path.

As soon as this force is removed, the object is "free" to leave the circular trajectory and continue its motion following a straight path at constant velocity, according to Newton's first law.

11. B) The acceleration will become 1/4 as much.

Re-arranging Newton's second law formula,

$F=ma \rightarrow a = \frac{F}{m}$ (1)

The force exerted on the satellite is actually the force of gravity:

$F=\frac{Gm M}{d^2}$ (2)

Substituting (2) into (1),

$a=\frac{F}{m}=\frac{GM}{d^2}$

We see that there is no dependance on the mass of the satellite (m), but only on the distance. Since in this problem the distance is doubled (d'=2d), the new acceleration will be:

$a'=\frac{GM}{(2d)^2}=\frac{1}{4}(\frac{GM}{d^2})=\frac{a}{4}$