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2 years ago

Answer and I will give you brainiliest

Physics

2 answers:

Snezhnost [94]2 years ago

6 0

A statement of what someone thinks will happen

svp [43]2 years ago

3 0

The answer to the question is C

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17. The four points built into the camera to ensure that each exposure can be oriented

3241004551 [841]

The diameter of the circle is 18 m. Eugene incorrectly says that the circumference of the circle is about 113.04 m. What mistake did Eugene make? Use 3.14 for pi.

7 0

2 years ago

The radius of Earth is about 6450 km. A 7070 N spacecraft travels away from Earth. What is the weight of the spacecraft at a hei

Triss [41]

Answer:

(a) 1767.43 N

(b) 182.45 N

Explanation:

Radius of earth, R = 6450 km

Weight of person, W = 7070 N

mass of person, m = W / g = 7070 / 9.8 = 721.4 kg

(a) h = 6450 km

The value of acceleration due to gravity on height is given by

$g' = g\left ( \frac{R}{R+h} \right )^2$

$g' = g\left ( \frac{6450}{6450+6450} \right )^2$

g' = g / 4 = 9.8 / 4 = 2.45 m/s^2

The weight of the person at such height is

W' = m x g' = 721.4 x 2.45

W' = 1767.43 N

(b) h = 33700 km

The value of acceleration due to gravity on height is given by

$g' = g\left ( \frac{R}{R+h} \right )^2$

$g' = g\left ( \frac{6450}{6450+33700} \right )^2$

g' = g x 0.0258 = 9.8 x 0.0258 = 0.253 m/s^2

The weight of the person at such height is

W' = m x g'

W' = 721.4 x 0.253

W' = 182.45 N

3 0

2 years ago

N experiment is performed in deep space with two uniform spheres, one with mass 27.0 and the other with mass 107.0 . They have e

Reptile [31]

Answer:

Explanation:

Apply the law of conservation of energy

$KE_i+PE_i=KE_f+PE_f$

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

from the law of conservation of the linear momentum

$m_1v_1=m_2v_2$

Therefore,

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

$=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]$

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

Substitute the values in the above result

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

$=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s$

B) the speed of the sphere with mass 107.0 kg is

$v_2=\frac{m_1v_1}{m_2}$

$=[\frac{27}{107} ](1.2664 \times 10^-^5)\\\\=3.195\times 10^-^6m/s$

C) the magnitude of the relative velocity with which one sphere is

$v_r=v_1+v_2\\\\=1.2664\times 10^-^5+3.195\times10^-^6\\\\=15.859\times10^-^6m/s$

D) the distance of the centre is proportional to the acceleration

$\frac{x_1}{x_2} =\frac{a_1}{a_2} \\\\=\frac{m_2}{m_1} \\\\=3.962$

Thus,

$x_1=3.962x_2$

and

$x_2=0.252x_1$

When the sphere make contact with eachother

Therefore,

$x_1+x_2+2r=41\\x_1+0,252x_1+2r=41\\1.252x_1+2r=41\\x_1=32.747-1.597r$

And

$x_1+x_2+2r=41\\3.962x_2+x_2+2r+41\\4.962x_2+2r=41\\x_2=8.262-0.403r$

The point of contact of the sphere is

$32.747-1.597r=8.262-0.403r\\\\r=\frac{24.485}{1.194} \\\\=20.506m$

3 0

2 years ago

What is the force required to lift the balloon with

Olin [163]

No force is required to lift that balloon. In fact, force is required to hold it down, and if you let go, it's up, up, and away.

Since the balloon's density is less than the density of the air around it, it's lighter than the air it displaces, there is a net upward buoyant force acting on it, and it floats up !

3 0

3 years ago

Anyone tell me what is the physics

Katyanochek1 [597]

Answer:

physic is branch of science in which math is its brother and will deal with the equation,law and evidence of natural phenomenon.

6 0

2 years ago