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lianna [129]
3 years ago
9

Why are force fields necessary to describe gravity?

Physics
2 answers:
artcher [175]3 years ago
8 0
Gravity is a non-contact force.
Vadim26 [7]3 years ago
5 0

Answer:

D. gravity is a non-contact force ​      

Explanation:

Gravitational force acts between two bodies having mass. The force is attractive in nature and is independent of the mass of the bodies. Thus, gravity can pull objects towards each other.

The magnitude of this force depends on the mass of the two bodies and inversely proportional to the square of the distance between them.

It is a non-contact force. So, to describe gravity, force fields are necessary. Force fields are used to describe non-contact forces- force on any body at any point in space.

Thus, correct option is D.  

You might be interested in
A charge of 35.0 μC is placed on conducting sphere A of radius 8.00 cm. Another identical conducting sphere B (radius 8.00 cm) c
Wewaii [24]

Answer:

a) 50μC

b) 37.45 m/s

Explanation:

a) If the spheres are connected the charge in both spheres tends to be equal. This because is the situation of minimum energy.

Thus, you have:

Q_T=35\mu C+65\mu C=100\mu C\\\\Q_s=\frac{Q_T}{2}=50\mu C

Hence, each sphere has a charge of 50μC.

b) You use the fact that the total work done by the electric force is equal to the change in the kinetic energy of the sphere. Then, you use the following equations:

\Delta W=\Delta K\\\\\int_{0.4}^\infty Fdr=\frac{1}{2}m[v^2-v_o^2]\\\\F=k\frac{Q^2}{r^2}\\\\v_o=0m/s\\\\m=0.08kg\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=kQ^2[-\frac{1}{r}]_{0.4}^{\infty}=\frac{kQ^2}{0.4m}=\frac{(8.98*10^9Nm^2/C^2)(50*10^{-6}C)^2}{0.4m}\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=56.125J

where you have used the Coulomb constant = 8.98*10^9 Nm^2/C^2

Next, you equal the total work to the change in K:

\frac{1}{2}mv^2=56.125J\\\\v=\sqrt{\frac{2(56.125J)}{m}}=\sqrt{\frac{2(56.125J)}{0.08kg}}=37.45\frac{m}{s}

hence, the speed of the spheres is 37.45 m/s

8 0
3 years ago
Why cant your rocket could never reach the speed of light?
Oksanka [162]

Answer:

The length of the object would shrink to zero which is not possible.

Explanation:

A rocket or any body cannot reach the speed of light because according to theory of relativity the and the Lorentz factor the length of the object would shrink to zero and the time dilation for that body would be infinite.

The Lorentz factor is given as:

\gamma=\frac{1}{\sqrt{\frac{v^2}{c^2} } }

where:

v = speed of the moving object

c = speed of light

4 0
3 years ago
una onda longitudinal tiene una frecuencia de 200 hz y una longitud de onda de 4.2m ¿cual es la rapidez de la onda?​
swat32

Answer:

v = 8.4 m/s

Explanation:

The question ays, "A longitudinal wave has a frequency of 200 Hz and a wavelength of 4.2m. What is the speed of the wave?".

Frequency of a wave, f = 200 Hz

Wavelength = 4.2 cm = 0.042 m

We need to find the speed of the wave. The formula for the speed of a wave is given by :

v=f\lambda\\\\v=200\times 0.042\\\\=8.4\ m/s

So, the speed of the wave is equal to 8.4 m/s.

4 0
3 years ago
When you lift a bowling ball with a force of 61.1 N, the ball accelerates upward with an acceleration a. If you lift with a forc
gtnhenbr [62]

Answer:

the weight of the ball is w = 51.94 N ( mass = 5.3 kg)

Explanation:

Following Newton's second law:

net force = mass * acceleration = weight/gravity * acceleration

then denoting 1 and 2 as the first and second lift

F₁ - w= w/g *a₁

F₂ -w = w/g *a₂ = w/g * 2.07a

dividing both equations

(F₂- w)/(F₁ -w)= 2.07

(F₂- w) = 2.07 * (F₁ -w)

1.07*w = 2.07*F₁ - F₂

w = (2.07*F₁ - F₂ )/ 1.07

replacing values

w = (2.07*61.1 N - 70.9 N )/ 1.07  = 51.94 N

then the weight of the ball is w = 51.94 N ( mass = 5.3 kg)

7 0
3 years ago
A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)
Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is U  =  \int\limits^T_0 {P(t)} \, dt

Compute U if the bulb remains on for 5h

Answer:

The value is  U  =  7.563 *10^{5} \  J

Explanation:

From the question we are told that

   The power rating of the bulb is P  =  100 \  W

   The resistance is   R =  143 \ \Omega

   The  voltage is  V  =  V_o  sin [2 \pi ft]

   The  energy expanded is U  =  \int\limits^T_0 {P(t)} \, dt

   The  voltage  V_o  =  110 \  V

   The frequency is  f =  60 \  Hz

    The  time considered is  t =  5 \  h  =  18000 \  s

Generally power is mathematically represented as

             P =  \frac{V^2}{ R}

=>          P =  \frac{( 110  sin [2 \pi * 60t])^2}{ 144}

=>           P =  \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}

So  

     U  =  \int\limits^T_0 { \frac{ 110^2*  [sin [120 \pi t])^2}{ 144}} \, dt

=>  U  =  \frac{110^2}{144} \int\limits^T_0 { (   sin^2 [120 \pi t]} \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | T} \atop {0}} \right.

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | 18000} \atop {0}} \right.

U =  \frac{110^2}{144} [\frac{18000}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi (18000))}{240 \pi} ] ]

=>   U  =  7.563 *10^{5} \  J

7 0
3 years ago
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