All categories

2 years ago

the gravitational force between the sun and earth is larger than sun and the moon. what do you think the reason for this is?

Physics

2 answers:

Tems11 [23]2 years ago

7 0

Answer: The greater an object's mass, the more gravitational force it exerts.

Explanation: So, to begin answering your question, Earth has a greater gravitational pull than the moon simply because the Earth is more massive. Sorry if I get this wrong. I am in 5th grade! ♥

ivann1987 [24]2 years ago

4 0

hi i need help with one of my question but i don’t know how else to contact you :/. anyway i posted the question and i added the text so i was wondering if u could help me !

You might be interested in

How much current is in a circuit that includes a 9-volt battery with a resistance of 3

Darya [45]

Just remember
Voltage = current times resistance

current = voltage over resistance

Current = 9/3 = 3

7 0

3 years ago

a 25 kg object is pushed with a horizontal force of 5 N esat across a table. If the force of friction is 2 N, what is the accele

Julli [10]

Answer:

Apply Newton's second law in the moving direction.

Explanation:

$\begin{aligned}F &= ma\\5N-2N &= 25kg\times a\\a & = \frac{3N}{25kg}\\&= 0.12ms^{-2}\end{aligned}$

Friction force applies in the opposite direction of motion; as a restriction.

8 0

3 years ago

Bernoulli's principle is responsible for most of the lift produced by an airplane wing.

Rina8888 [55]

Answer:

huiiiiuuu beautiful...

5 0

2 years ago

An alpha particle (Î±), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an elec

vaieri [72.5K]

Answer:

Speed = 575 m/s

Mechanical energy is conserved in electrostatic, magnetic and gravitational forces.

Explanation:

Given :

Potential difference, U = $-3.45 \times 10^{-3} \ V$

Mass of the alpha particle, $m_{\alpha} = 6.68 \times 10^{-27} \ kg$

Charge of the alpha particle is, $q_{\alpha} = 3.20 \times 10^{-19} \ C$

So the potential difference for the alpha particle when it is accelerated through the potential difference is

$U=\Delta Vq_{\alpha}$

And the kinetic energy gained by the alpha particle is

$K.E. =\frac{1}{2}m_{\alpha}v_{\alpha}^2$

From the law of conservation of energy, we get

$K.E. = U$

$\frac{1}{2}m_{\alpha}v_{\alpha}^2 = \Delta V q_{\alpha}$

$v_{\alpha} = \sqrt{\frac{2 \Delta V q_{\alpha}}{m_{\alpha}}}$

$v_{\alpha} = \sqrt{\frac{2(3.45 \times 10^{-3 })(3.2 \times 10^{-19})}{6.68 \times 10^{-27}}}$

$v_{\alpha} \approx 575 \ m/s$

The mechanical energy is conserved in the presence of the following conservative forces :

-- electrostatic forces

-- magnetic forces

-- gravitational forces

5 0

3 years ago

The four-wheel-drive all-terrain vehicle has a mass of 385 kg with center of mass G2. The driver has a mass of 75 kg with center

jenyasd209 [6]

The coefficient of friction is missing and it has a value of μ = 0.4

Answer:

a = 3.924 m/s²

Explanation:

I've attached the kinematic free body diagram.

Taking the sum of all upward and downward forces,

EFy = 0;

N1 + N2 - m_p•g - m_v•g = 0

N1 + N2 = m_p•g + m_v•g

Where;

N1 and N2 are the normal reactions at the wheels

m_p is the mass of the driver

m_v is the mass of the vehicle

g is the acceleration due to gravity.

Plugging in the relevant values in the question,we obtain;

N1 + N2 = (385 + 75) x 9.81

N1 + N2 = 4512.6N - - - (eq1)

Now, taking sum of all horizontal forces;

EFx = (m_p + m_v) x a

So,

μ(N1 + N2) = (mp + mv) x a

Thus,

0.4(N1 + N2) = (385 + 75)a

0.4(N1 + N2) = 460a

N1 + N2 = 1150a

From eq(1),N1 + N2 = 4512.6N

Thus,

1150a = 4512.6N

a = 4512.6/1150

a = 3.924 m/s²

Therefore, the acceleration, a = 3.924 m/s²

7 0

3 years ago