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3 years ago

5

Consider the roller coaster on the track. If you want to increase the kinetic energy of the roller coaster as it proceeds throug

h the course, what is the FIRST thing you should do?
A) Increase the height at point A.
B) Decrease the height at point A.
C) Increase the height at point B.
D) Decrease the height at point B.

1 answer:

a_sh-v [17]3 years ago

5 0

Answer:

The answer is A and D but if you want one anwer i think the anwer is A

Explanation:

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A planet exerts a gravitational force of magnitude 9e22 N on a star. If the planet were 2 times closer to the star (that is, if

Dmitrij [34]

To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.

Planet gravitational force

$F_p = 6*10^{22}N$

$F_p = \frac{GMm}{R^2}$

$F_p = 9*10^{22}N$

Distance between planet and star

$r = \frac{R}{2}$

Gravitational force is

$F = \frac{GMm}{r^2}$

Applying the new distance,

$F = \frac{GMm}{(\frac{R}{2})^2}$

$F = 4\frac{GMm}{R^2}$

Replacing with the previous force,

$F = 4F_p$

Replacing our values

$F= 4(9*10^{22}N)$

$F = 36*10^{22}N$

Therefore the magnitude of the force on the star due to the planet is $36*10^{22}N$

5 0

3 years ago

A convex thin lens with refractive index of 1.50 has a focal length of 30cm in air. When immersed in a certain transparent liqui

GalinKa [24]

Answer:

$n_l = 1.97$

Explanation:

given data:

refractive index of lens 1.50

focal length in air is 30 cm

focal length in water is -188 cm

Focal length of lens is given as

$\frac{1}{f} =\frac{n_2 -n_1}{n_1} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

$\frac{1}{f} =\frac{n_{g} -n_{air}}{n_{air}} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

$\frac{1}{f} =\frac{n_{g} -1}{1} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

focal length of lens in liquid is

$\frac{1}{f} =\frac{n_{g} -n_{l}}{n_{l}} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

$=\frac{n_{g} -n_{l}}{n_{l}} [\frac{1}{(n_{g} - 1) f}$

rearrange fro $n_l$

$n_l = \frac{n_g f_l}{f_l+f(n_g-1)}$

$n_l = \frac{1.50*(-188)}{-188 + 30(1.50 -1)}$

$n_l = 1.97$

7 0

3 years ago

Help! Don’t understand any of this!!

Kryger [21]

Yea me either. Good look

5 0

3 years ago

70-kg man standing on a scale in an elevator notes that as the elevator rises, the scale reads 844 N. What is the acceleration o

mario62 [17]

Answer : The acceleration of the elevator is, $2.25m/s^2$

Explanation :

Formula used to calculate the acceleration of the elevator is:

$F=mg+ma\\\\ma=F-mg\\\\a=\frac{F-mg}{m}$

where,

F = force = 844 M

m = mass of man = 70 kg

g = acceleration due to gravity = $9.8m/s^2$

a = acceleration of the elevator = ?

Now put all the given values in the above formula, we get:

$a=\frac{F-mg}{m}$

$a=\frac{844N-(70kg\times 9.8m/s^2)}{70kg}$

$a=2.25m/s^2$

Thus, the acceleration of the elevator is, $2.25m/s^2$

4 0

3 years ago

4. What is the electric field strength (E) at a distance of 0.50 m from a 1.00×10°C charge?

Romashka-Z-Leto [24]

Answer:

3.6*10¹⁰N/C

Explanation:

The formula for Electric field strength is expressed as:

E = kQ/r²

k is the coulombs constant = 9*10^9Nm²/C²

Q is the charge = 1.00×10°C

r is the distance = 0.50m

Substitute the parameters into the formula as shown:

E = kQ/r²

E = 9*10^9(1)/0.5²

E = 9*10^9/0.25

E = 36*10^9

<em>E = 3.6*10¹⁰N/C</em>

<em>Hence the electric field strength is 3.6*10¹⁰N/C </em>

3 0

3 years ago