. Which is the transfer of energy as waves moving through space?

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 speedboat moving at 28 m/s approaches a no-wake buoy marker 91 m ahead. The pilot slows the boat with a constant acceleration

Tpy6a [65]

Answer:

7.5 m/s

Explanation:

We can find its velocity when it reaches the buoy by applying one of Newton's equations of motion:

$v^2 = u^2 + 2as$

where v = final velocity

u = initial velocity

a = acceleration

s = distance traveled

From the question:

u = 28 m/s

a = -4 $m/s^2$

s = 91 m

Therefore:

$v^2 = 28^2 + 2 * (-4) * 91\\\\v^2 = 784 + -728 = 56\\\\v = \sqrt{56}\\ \\v = 7.5 m/s$

The velocity of the boat when it reaches the buoy is 7.5 m/s.

6 0

3 years ago

To determine power, it is necessary to know the A. energy required and the time it takes. B. amount of work and the distance cov

Y_Kistochka [10]

The answer for this would be A. since power is Joules/seconds and energy is rated in Joules

5 0

3 years ago

Read 2 more answers

Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slid

nataly862011 [7]

Answer:

A)

the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B)

data that were not necessary to the solution are;

a) mass of truck and b) mass of load

Explanation:

Given that;

mass of load $m_{LS}$ = 10000 kg

mass of flat bed $m_{FB}$ = 20000 kg

initial speed of truck $v_{0}$ = 12 m/s

coefficient of friction between the load sits and flat bed μs = 0.5

A) the minimum stopping distance for which the load will not slide forward relative to the truck.

Now, using the expression

Fs,max = μs $F_{N}$ -------------let this be equation 1

where $F_{N}$ = normal force = mg

so

Fs,max = μs mg

ma $_{max}$ = μs mg

divide through by mass

a $_{max}$ = μs g ---------- let this be equation 2

in equation 2, we substitute in our values

a $_{max}$ = 0.5 × 9.8 m/s²

a $_{max}$ = 4.9 m/s²

now, from the third equation of motion

v² = u² + 2as

$v_{f}$ ² = $v_{0}$ ² + 2aΔx

where $v_{f}$ is final velocity ( 0 m/s )

a is acceleration( - 4.9 m/s² )

so we substitute

(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx

0 = 144 m²/s² - 9.8 m/s²Δx

9.8 m/s²Δx = 144 m²/s²

Δx = 144 m²/s² / 9.8 m/s²

Δx = 14 m

Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B) data that were not necessary to the solution are;

a) mass of truck and b) mass of load

3 0

3 years ago

Suppose a car traveling at 8m/s is brought to rest in a distance of 20m.what is it's deceleration and time taken.

LiRa [457]

Answer:

Time - taken = 2.5 s

deceleration= -8 m/s²

Solution:

Given:

speed, v = 8 m/s

distance, d = 20m

To Find:

deacceleration = ?

As we know speed is defined as

v = d/t

plugging in the values

t = 20/ 8

t = 2.5s

Now from deceleration formula

a = - v/ t

a = - 20/ 2.5

a = - 8 m/s²

Thus, the time taken and acceleration is 2.5 s and -8 m/s²

respectively.

Learn more about deceleration here:

brainly.com/question/13354629

#SPJ4

7 0

2 years ago