If a 50.0-kg mass weighs 554 n on the planet saturn, calculate saturn’s radius

Answer:

Explanation:

To begin with, we must determine the pressure acting on the sphere, which is calculated using:

where

denotes the atmospheric pressure

represents the density of the water

signifies the acceleration due to gravity

indicates the depth

By substituting these values,

The sphere's radius is calculated as r = d/2 = 1.1 m/2 = 0.55 m

Thus, the sphere's total surface area can be expressed as

Consequently, the inward force acting on the sphere equals

Answer:

7.166 hours = 430 minutes.

Explanation:

As both trains are approaching each other on the same track, their relative speed is the sum of their individual speeds. Hence, the time until they intersect (and inevitably collide) is determined by how long it takes for speeds of 65 mph and 55 mph to cover the total distance of 860 miles. One train will cover part of the distance, while the other will cover the remainder. To calculate the required time, we can apply the formula:

1 hour ---> 120 miles

X ----> 860 miles; hence X = (860 miles * 1 hour)/120 miles = 43/6 hours = 7.16666 hours. To convert this into minutes, recall that 1 hour equals 60 minutes; therefore, 43/6 hours * 60 minutes/hour = 430 minutes.

Answer:

Explanation:

a) La fuerza neta que actúa sobre la caja en la dirección vertical es:

Fnet=Fg−f−Fp *sin45 °

aquí Fg representa la fuerza gravitacional, f es la fuerza de fricción, y Fp es la fuerza de empuje.

Fnet=ma

ma=Fg−f−Fp *sin45 °

​a=

=0.24 m/s²

Vf =Vi +at

=0.48+0.24*2

Vf=2.98 m/s

b)

Fnet=Fg−f−Fp *sin45 °

=Fg−0.516Fp−Fp *sin45 °

=30-1.273Fp

Fnet=0 (Ya que la velocidad es constante)

Fp=30/1.273

=23.56 N

A. The horizontal component of velocity is
vx = dx/dt = π - 4πsin(4πt + π/2)
vx = π - 4πsin(0 + π/2)
vx = π - 4π(1)
vx = -3π

b. vy = 4πcos(4πt + π/2)
vy = 0

c. m = sin(4πt + π/2) / [πt + cos(4πt + π/2)]

d. m = sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]

e. t = -1.0

f. t = -0.35

g. To find t, set
vx = π - 4πsin(4πt + π/2) = 0
Then use this to calculate vxmax

h. To determine t, set
vy = 4πcos(4πt + π/2) = 0
Then use this to find vymax

i. s(t) = [x(t)^2 + y(t)^2]^(1/2)

h. s'(t) = d[x(t)^2 + y(t)^2]^(1/2) / dt

k and l. Determine the values for t
d[x(t)^2 + y(t)^2]^(1/2) / dt = 0
And substitute to find both the maximum and minimum speeds.

The question lacks details. Here is the full question.

The accompanying image was captured with a camera capable of shooting between one and two frames per second. A series of photos was merged into this single image, meaning the vehicles depicted are actually the same car, documented at different intervals.

Assuming the camera produced 1.3 frames per second for this image and that the length of the car is approximately 5.3 meters, based on this information and the photo, how fast was the car moving?

Answer: v = 6.5 m/s

Explanation: The problem requires calculating the car's velocity. Velocity can be computed using:

Since the camera captured 7 images of the car and its length is noted as 5.3, the car's displacement is:

Δx = 7(5.3)

Δx = 37.1 m

The camera operates at 1.3 frames per second and recorded 7 images, thus the time driven by the car is:

1.3 frames = 1 s

7 frames = Δt

Δt = 5.4 s

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v = 6.87 m/s

<pthe car="" moved="" at="" an="" estimated="">velocity of 6.87 m/s.

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