Answer:
Angle is 55.52°
and Initial Speed is v=26.48 m/s
Explanation:
Given data

Applying the kinematics equations for motion with uniform acceleration in x and y direction
So

Put the value of v₀ from equation (1) to equation (2)
So

Put that angle in equation (1) or equation (2) to find the initial velocity
So from equation (1)

<h2>
Answer: 56.718 min</h2>
Explanation:
According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.
</em>
In other words, this law states a relation between the orbital period
of a body (moon, planet, satellite) orbiting a greater body in space with the size
of its orbit.
This Law is originally expressed as follows:
(1)
Where;
is the Gravitational Constant and its value is
is the mass of Mars
is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)
If we want to find the period, we have to express equation (1) as written below and substitute all the values:
(2)
(3)
(4)
Finally:
This is the orbital period of a spacecraft in a low orbit near the surface of mars
Answer:
213 nA
2.13 mA
851e^-t μA
Explanation:
We have a pretty straightforward question here.
Ohms Law states that the current in an electric circuit is directly proportional to the voltage and inversely proportional to the resistance in the circuit. It is mathematically written as
V = IR, since we need I, we can write that
I = V/R
a) at V = 1 mV
I = (1 * 10^-3) / 4.7 * 10^3
I = 2.13 * 10^-7 A or 213 nA
b) at V = 10 V
I = 10 / 4.7 * 10^3
I = 0.00213 A or 2.13 mA
c) at V = 4e^-t
I = 4e^-t / 4.7 * 10^3
I = 0.000851e^-t A or 851e^-t μA
(1) The image of an object placed further from the lens than the focal point will be upside down and smaller than the object.
(2) When light rays reflect, they bounce back.
(3) Images formed by a concave lens will look magnified.
(4) When light rays enter a different medium, they bend.
<h3>
1.0 Object placed further from the lens than the focal point</h3>
The image of an object placed further from the lens than the focal point will be diminished and inverted.
Thus, the correct answer will be "upside down and smaller than the object".
<h3>2.0 What is reflection of light?</h3>
The ability of light to bounce back when it strike a hard surface is known as refection.
<h3>3.0 Image formed by concave lens</h3>
A concave lens is diverging lens is usually virtual, erect and magnified.
<h3>4.0 Refraction of light</h3>
The change in speed of light when it travels from medium to another medium is known as refraction. Refraction is also, the ability of light to bend around obstacles.
Learn more about reflection and refraction of light here: brainly.com/question/1191238
Answer:
As the cars ascend the next hill, some kinetic energy is transformed back into potential energy. Then, when the cars descend this hill, potential energy is again changed to kinetic energy. This conversion between potential and kinetic energy continues throughout the ride.
Explanation:
hope it helps U