Answer:
The answer to your question is: a = 2 m/s²
Explanation:
I' ll solve it in this way, tell me if I'm right. Because we need to consider that in the moon there is no gravity.
Data
F = 50 N
a = 4.0 m/s²
Moon
F = 25 N
a = ?
Formula
F = m x a
First, calculate the mass
m = F / a
m = 50 / 4
m = 12.5 kg
Now, calculate the acceleration
a = F / m
a = 25 / 12.5
a = 2 m/s²
<span>We know that v = u + at
He stops when the speed is zero; therefore
0 = 15 - 3.35 t
t = 4.48
It takes 4.48 seconds after the driver turns on the brake for him to stop.
We know
s = ut + (1/2)at</span>²<span>
s = 15*4.48 - *(1/2)(3.25)(4.48)</span>²<span>
s = 33.4 meters
The distance he travels before applying the brakes is:
s = 0.21 * 15
s = 3.15 m
The total distance he travels is:
3.15 + </span>33.4 = 36.55 meters
He will be able to stop in time.
Answer:
98.4 N
Explanation:
Given that the body weighs 800 N on earth.
Thus,
Weight = mass x acceleration due gravity
i.e W = mg
800 = m x 10
m =
= 80 kg
The mass of the body is 80 kg.
To be able to determine its weight on the planet, we have to first calculate the gravitational pull of the planet.
But,
g =
Where: G is the Newton's universal gravitation, M is the mass of the planet and r is the radius of the planet.
g =
=
= 1.2275
g = 1.23 m/
Thus,
Weight of the body on the planet = mg
= 80 x 1.23
= 98.4 N
The weight of the body on the planet is 98.4 N
Answer:
q_enclosed = 759.57 nC
Explanation:
A cylindrical shell of radius 7.00 cm and length 2.48 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 15.3 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C.
Find the value of the net charge on the shell.
Given that Length L of the cylinder shell is 248 cm and the diameter d is 7.0 cm. We can approximate L >> d. This approximation concludes that the cylindrical shell is infinitely long, when we are examining the Electric fiel strength E from the ends of the shell.
Hence,
E = q_enclosed / 2*pi*ε_o *r*l
Where r = d / 2, then
q_enclosed = E*(2*π*r*l*ε_o)
q_enclosed = (36000)*(2*π*8.85*10^-12 *0.153*2.48)
q_enclosed = 759.57 nC
Answer:
(c) The planet must have a mass about the same as the mass of Jupiter,
(d) The planet must be closer to the star than Earth is to the Sun.
Explanation:
Astrometry is the ideal method to detect high-mass planets that are close to their star. That is because the gravitational effect that it will have the planet over its host star will be greater. This effect can be seen as a wobble in the star as a consequence of how they orbit a common center of mass¹. The center of mass will be closer to the most massive object, So, in the case of an extrasolar planet with masses like Jupiter (Jovian), this point will be a little bit farther from the star, making the wobble more notable than in a system with a low-mass planet.
Key terms:
Astrometry: study of the position of the stars over time in the sky.
¹Center of mass: a geometrical point in which the mass from a whole system is summed.