All categories

3 years ago

What forces act upon the moon as it orbits the earth

Physics

2 answers:

ValentinkaMS [17]3 years ago

7 0

the axis acts against and it would be a contact force

Mashcka [7]3 years ago

5 0

The correct answer would be gravitational force.

Hope this helped :)

You might be interested in

1. What causes velocity to change?

tresset_1 [31]

The amount of force an object has will change the velocity

4 0

3 years ago

Read 2 more answers

A scientist records the motion of an object. She writes down that the

Phantasy [73]

The answer is Velocity

3 0

3 years ago

A zero-order reaction has a constant rate of 2.30×10−4 M/s. If after 80.0 seconds the concentration has dropped to 1.50×10−2 M,

dolphi86 [110]

Answer:

Initial concentration of the reactant = 3.34 × 10^(-2)M

Explanation:

Rate of reaction = 2.30×10−4 M/s,

Time of reaction = 80s

Final concentration = 1.50×10−2 M

Initial concentration = Rate of reaction × Time of reaction + Final concentration

= 2.30×10−4 M/s × 80s + 1.50×10−2 M = 3.34 × 10^(-2)M

Initial concentration = 3.34 × 10^(-2)M

6 0

3 years ago

Oscilloscopes have parallel metal plates inside them to deflect the electron beam. These plates are called the deflecting plates

Alina [70]

Answer:

The capacitance of the deflecting plates is $C=1.59 pF$ .

Explanation:

The expression for the capacitance of the capacitor in terms of area and distance is as follows;

$C=\frac{\varepsilon _{0}A}{d}$

Here, C is the capacitance, A is the area, d is the distance and $\varepsilon _{0}$ is the absolute permittivity.

Convert the side of the square from cm to m.

s= 3.0 cm

s= 0.030 m

Calculate the area of the square.

$A= s_^{2}$

Put s= 0.030 m.

$A=(0.030)_^{2}$

$A=9\times10^{-4}m^{-2}$

Convert distance from mm to m.

d= 5.0 mm

$d=5\times10^{-3}m$

Calculate the capacitance of the deflecting plates.

$C=\frac{\varepsilon _{0}A}{d}$

Put $d=5\times10^{-3}m$ , $A=9\times10^{-4}m^{-2}$ and $\varepsilon _{0}=8.85\times 10^{-12}Fm^{-1}$ .

$C=\frac{(8.85\times 10^{-12})(9\times10^{-4})}{5\times10^{-3}}$

$C=1.59\times 10^{-12}F$

$C=1.59 pF$

Therefore, the capacitance of the deflecting plates is $C=1.59 pF$ .

4 0

3 years ago

A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She th

Andre45 [30]

The horizontal distance covered by the ball before hitting the water is 70.4 m

Explanation:

The motion of the ball is the motion of a projectile, so it consists of two independent motions:

A uniform motion along the horizontal (x) direction
A uniformly accelerated motion along the vertical (y) direction

We start by calculating the time of flight of the ball. This can be done by analyzing the vertical motion. We can use the following suvat equation:

$s=u_y t + \frac{1}{2}at^2$

where:

s = -16.5 m is the vertical displacement of the ball (it is negative because we take upward as positive direction)

$u_y$ is the initial vertical velocity of the ball, which is given by

$u_y = u sin \theta$

where

u = 23.5 m/s is the initial velocity

$\theta=33.5^{\circ}$ is the angle of projection

Substituting,

$u_y=(23.5)(sin 33.5^{\circ})=13.0 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity, downward

Substituting everything into the equation we get:

$-16.5=13.0t-4.9t^2\\4.9t^2-13.0t-16.5=0$

Solving the equation for t, we find the time of flight of the ball:

t = -0.94 s

t = 3.59 s

We ignore the 1st solution since it is negative, so the ball reaches the water after 3.59 seconds.

Now we analyze the horizontal motion of the ball. The horizontal velocity is constant and it is:

$v_x=u cos \theta=(23.5)(cos 33.5^{\circ})=19.6 m/s$

Therefore, the horizontal distance covered in a time t is

$d=v_x t$

And substituting t = 3.59 s, we find

$d=(19.6)(3.59)=70.4 m$

So, the horizontal distance covered by the ball before hitting the water is 70.4 m.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

4 0

3 years ago