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andrezito [222]
3 years ago
14

If a metal ball suspended by a rod is at rest, which force is responsible for balancing the force due to gravity?

Physics
2 answers:
sineoko [7]3 years ago
4 0
The normal force applied by the rod on tbe metal ball is the force balancing the force due to gravity.

The normal force is the force that keeps any object from going through another object and that is why tables and floors can keep things up.
cestrela7 [59]3 years ago
3 0
Any force coming from the surface and acting at a right angle to the surface is called the Normal Force<span>.
normal force </span><span>is responsible for balancing the force due to gravity.</span>
You might be interested in
An infinitely long line of charge has a linear charge density of 7.50×10^−12 C/m . A proton is at distance 14.5 cm from the line
Nata [24]

Answer:

10.22 cm

Explanation:

linear charge density, λ = 7.5 x 10^-12 C/m

distance from line, r = 14.5 cm = 0.145 m

initial speed, u = 3000 m/s

final speed, v = 0 m/s

charge on proton, q = 1.6 x 10^-19 C

mass of proton, m = 1.67 x 10^-27 kg

Let the closest distance of proton is r'.

The potential due t a line charge at a distance r' is given  by

V=-2K\lambda ln\left (\frac{r'}{r}  \right )

where, K = 9 x 10^9 Nm^2/C^2

W = q V

By use of work energy theorem

Work = change in kinetic energy

qV = 0.5m(u^{2}-v^{2})

By substituting the values, we get

V=\frac{mu^{2}}{2q}

-2K\lambda ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{2q}

- ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{4Kq\lambda }

- ln\left ( \frac{r'}{r} \right )=\frac{1.67 \times 10^{-27}\times 3000\times 3000}{4\times 9\times 10^{9}\times 1.6\times 10^{-19}\times 7.5\times 10^{-12} }

- ln\left ( \frac{r'}{r} \right )=0.35

\frac{r'}{r} =e^{-0.35}

\frac{r'}{r} =0.7047

r' = 14.5 x 0.7047 = 10.22 cm

5 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}

12. C) divide: distance : velocity

Velocity is distance divided by time:

v=\frac{d}{t}

Multiplying by t on both sides,

v\cdot t = \frac{d}{t} \cdot t \rightarrow vt = d

and dividing by v on both sides,

\frac{vt}{v}=\frac{d}{v} \rightarrow t = \frac{d}{v}

13. A) -8 km/h

Let's call:

v_k = +5 km/h the velocity of the kite

v_p = -3 km/h your velocity (opposite direction)

In your frame of reference, the velocity of the kite will be:

v'_k = v_p - v_k = -3 -(+5) = -8 km/h

3 0
4 years ago
Read 2 more answers
--------------is an aquatic animal which emits electric discharge.
Varvara68 [4.7K]

Answer:

There are 5 of 'em.

It's so amazing.

1. The electric eel; It can generate about 370-650 volts of electric current from it's body.

2. The electric catfish!

About 350-450volts of electric current.

3. The electric Ray.

Say no more. It can generate about 37-220volts.

4. The electric stargazer (really stargazer, lol, and you live down the sea)

About 50volts

and

5. Skate.

Just 4volts.

You know more than to be careful when you go diving in the sea.

I love your question btw. I learnt also.

4 0
3 years ago
What is the amount of heat required to turn 2kg of ice into water at 80°
mote1985 [20]

Explanation:

Sorry if I toke a while (needed to finish some work) but here you go!!! :D.

5 0
3 years ago
Matter is made up of very small particles
aivan3 [116]
<h2>Answer:</h2>

The given statement on matter is true.

<h2>Explanation:</h2>

The matter refers to any substance or object that occupies volume. A matter always has mass which are composed of huge number of microscopic particles named atoms. The atoms are comprises subatomic particle, however, it does not contain particle like photons (massless particle).

The matter is classified widely based on many factors and based on its existence it is classified as liquid solid, and fluid. Depending on the structure, the matter is classified into baryonic matter, hadronic matter, degenerate matter, and strange matter.  

Thus, the matter comprises tiny particles.

6 0
3 years ago
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