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4 years ago

Which force is greater the earth’s pull on the moon or the moon’s pull on the earth

Physics

1 answer:

Vitek1552 [10]4 years ago

7 0

Answer:

The earth's pull on the moon

Explanation:

Earth exerts a gravitational pull on the moon 80 times stronger than the moon's pull on the Earth.

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Multiple Intelligences :

vlada-n [284]

Answer:

all of the above

PLS MARK ME AS BRAINLIAST

8 0

3 years ago

2. A force of 50 N is applied to the object for a distance of 2.0 m. Assume that object（the mass of the object is 3kg）

Gwar [14]

Answer:

2. A force of 50 N is applied to the object for a distance of 2.0 m. Assume that object（the mass of the object is 3kg）

was at rest at the beginning, what speed did it achieved because of the work done on it? (Hint:

Calculate the works performed by the force first.)

I figured that is 8.2m/S，I am just not sure can anyone help me i much appreciate it.

4 0

3 years ago

The temperature coefficient of a certain conducting material is 5.74 × 10-3 (°C)-1. (a) At what temperature would the resistance

Alex777 [14]

Answer:

Temperature at which the resistance is twice the resistance at $20^{\circ}C$ is $194.216^{\circ}C$

Solution:

As per the question:

Temperature coefficient, $\alpha = 5.74\times 10^{- 3}^{\circ}C$

Reference temperature, $T_{o} = 20^{\circ}C$

Resistance, $R_{t} = 2R_{o}$

Now, using the formula:

$R_{t} = R_{o}(1 + \alpha \Delta T)$

$2R_{o} = R_{o}(1 + \alpha \times (T _ T_{o}))$

$2 = 1 + 5.74\times 10^{- 3}\times (T - 20^{\circ})$

$\frac{1}{5.74\times 10^{- 3}} = T - 20^{\circ}$

$T = 174.216 + 20 = 194.216^{\circ}$

Yes, this temperature holds for all all the conductors of copper, irrespective of the size and shape of the conductor.

5 0

3 years ago

Which missing item would complete this beta decay reaction?

Flauer [41]

For us to understand the missing item that would complete beta decay reaction, we need to achieve in depth understanding of chemical formulas and nuclear symbols. Next is to have great comprehension of the following points:
<span>1.) Neutron in nucleus breaks down and changes into a proton.
2) Then it emits an electron, as well as an anti-neutrino which go into space.
3) Lastly, atomic number continuously goes UP while mass number remains unchanged.</span>

8 0

3 years ago

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

spayn [35]

Answer:

$v_0 = 3.53~{\rm m/s}$

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

$x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}$

For the y-direction gives

$v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t$

Combining both equation yields the y_component of the final velocity

$v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}$

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

$\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}$

6 0

3 years ago