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4 years ago

the force is found by multiplying the mass of an object by velocity at which it travels. true or false

Physics

1 answer:

Katarina [22]4 years ago

4 0

False.

Force is found by multiplying mass of an object by its acceleration.

F = mass * acceleration

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A planet is 15 light years from earth. At which of the following speeds will the crew of a spaceship complete the trip to the pl

Svetradugi [14.3K]

Answer:

option c) 0.88c is the correct answer

Explanation:

using the Lorrentz equation we have

$t=\frac{d}{v}\sqrt{1-(\frac{v}{c})^2}$

where,

t = time taken to cover the distance

d = Distance

v = velocity

c = speed of light

given

d = 15 light years

Now,

$8years=\frac{15years\times c}{v}\sqrt{1-(\frac{v}{c})^2}$

$(\frac{v}{c})^2\times (\frac{8}{15})^2=1-(\frac{v}{c})^2$

$(\frac{v}{c})^2\times 0.284=1-(\frac{v}{c})^2$

$(\frac{v}{c})^2\times 0.284 + (\frac{v}{c})^2 =1$

$(1.284\times \frac{v}{c})^2 =1$

$(\frac{v}{c}) =\sqrt{\frac{1}{0.284}}$

$v=0.88c$

7 0

3 years ago

(a) What is the escape speed on a spherical asteroid whose radius is 500. km and whose gravitational acceleration at the surface

navik [9.2K]

Answer:

a) v= 1732.05m/s

b) d=250000m

c) v= 1414.214m/s

Explanation:

Notation

M= mass of the asteroid

m= mass of the particle moving upward

R= radius

v= escape speed

G= Universal constant

h= distance above the the surface

Part a

For this part we can use the principle of conservation of energy. for the begin the initial potential energy for the asteroid would be $U_i =-\frac{GMm}{R}$ .

The initial kinetic energy would be $\frac{1}{2}mv^2$ . The assumption here is that the particle escapes only if is infinetely far from the asteroid. And other assumption required is that the final potential and kinetic energy are both zero. Applying these we have:

$-\frac{GMm}{R}+\frac{1}{2}mv^2=0$ (1)

Dividing both sides by m and replacing $\frac{GM}{R}$ by $a_g R$

And the equation (1) becomes:

$-a_g R+\frac{1}{2} v^2=0$ (2)

If we solve for v we got this:

$v=\sqrt{2 a_g R}=\sqrt{2x3\frac{m}{s^2}x500000m}=1732.05m/s$

Part b

When we consider a particule at this surface at the starting point we have that:

$U_i=-\frac{GMm}{R}$

$K_i=\frac{1}{2}mv^2$

Considering that the particle is at a distance h above the surface and then stops we have that:

$U_f=-\frac{GMm}{R+h}$

$K_f=0$

And the balance of energy would be:

$-\frac{GMm}{R}+\frac{1}{2}mv^2 =-\frac{GMm}{R+h}$

Dividing again both sides by m and replacing $\frac{GM}{R}$ by $a_g R^2$ we got:

$-a_g R+\frac{1}{2}v^2 =-\frac{a_g R^2}{R+h}$

If we solve for h we can follow the following steps:

$R+h=-\frac{a_g R^2}{-a_g R+\frac{1}{2}v^2}$

And subtracting R on both sides and multiplying by 2 in the fraction part and reordering terms:

$h=\frac{2a_g R^2}{2a_g R-v^2}-R$

Replacing:

$h=\frac{2x3\frac{m}{s^2}(500000m)^2}{2(3\frac{m}{s^2})(500000m)-(1000m/s)^2}- 500000m=250000m$

Part c

For this part we assume that the particle is a distance h above the surface at the begin and start with 0 velocity so then:

$U_i=-\frac{GMm}{R+h}$

$K_i=0$

And after the particle reach the asteroid we have this:

$U_f=-\frac{GMm}{R}$

$K_f=\frac{1}{2}mv^2$

So the balance of energy would be:

$-\frac{GMm}{R+h}=-\frac{GMm}{R}+\frac{1}{2}mv^2$

Replacing again $a_g R^2$ instead of GM and dividing both sides by m we have:

$-\frac{a_g R^2}{R+h}=-a_g R+\frac{1}{2}v^2$

And solving for v:

$a_g R-\frac{a_g R^2}{R+h}=\frac{1}{2}v^2$

Multiplying both sides by two and taking square root:

$v=\sqrt{2a_g R-\frac{2a_g R^2}{R+h}}$

Replacing

$v=\sqrt{2(3\frac{m}{s^2})(500000m)-\frac{2(3\frac{m}{s^2}(500000m)^2}{500000+1000000m}}=1414.214m/s$

3 0

3 years ago

The plates of a parallel plate capacitor are separated by d = 1.6 cm. The potential of the negative plate is 0 V, and the potent

yaroslaw [1]

Answer:

The electric field between the plates is 1875 V/m.

Explanation:

Given that,

Separated d=1.6 cm

Potential of negative plate = 0 V

Potential halfway between the plates = 15 V

We need to calculate the total potential

Using formula of potential

$V=2\times\text{Potential halfway between the plates}$

Put the value into the formula

$V=2\times15$

$V=30\ V$

We need to calculate the electric field between the plates

Using formula of electric field

$E=\dfrac{V}{d}$

$E=\dfrac{30}{1.6\times10^{-2}}$

$E=1875\ V/m$

Direction is negative as the field always points from positive to negative plate

Hence, The electric field between the plates is 1875 V/m.

5 0

3 years ago

A cruise ship makes its way from one island to another. The ship is in motion compared with which reference point?

SVEN [57.7K]

As the ship is not in motion compared with passengers on the ship so
it will be
a lighthouse on a nearby island
so correct answer will be C
hope it helps

3 0

3 years ago

Explain what makes rhombus different from other quadrilaterals

navik [9.2K]

The critical difference is the fact that all four sides of a rhombus have
the same length.

There are several other distinguishing characteristics, but if you make the
lengths of all the sides equal, then the other things happen automatically.

6 0

3 years ago