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3 years ago

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Susan, driving north at 53 mphmph , and Shawn, driving east at 63 mphmph , are approaching an intersection. Part A What is Shawn

's speed relative to Susan's reference frame?

1 answer:

mafiozo [28]3 years ago

5 0

Answer:

Shawn's speed relative to Susan's speed = 10 mph

Resultant velocity = 82.32 mph

Explanation:

The given data :-

i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.

ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.

iii) The speed of both Susan and Shawn is relative to earth.

iv) The angle between Susan in north and Shawn in east is 90°.

We have to find Shawn's speed relative to Susan's speed.

v₂₁ = v₂ - v₁ = 63 - 53 = 10 mph

Resultant velocity,

$v = \sqrt{v_{2} ^{2}+ v_{1} ^{2} } =\sqrt{63^{2} +53^{2} }$

v = 82.32 mph

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A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

$M$ - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is:

$F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}$

$F = 6841.905\,N$

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0

3 years ago

A pendulum is made up of a small sphere of mass 0.500 kg attached to a string of length 0.950 m. The sphere is swinging back and

Semenov [28]

Answer:

W = 0.842 J

Explanation:

To solve this exercise we can use the relationship between work and kinetic energy

W = ΔK

In this case the kinetic energy at point A is zero since the system is stopped

W = K_f (1)

now let's use conservation of energy

starting point. Highest point A

Em₀ = U = m g h

Final point. Lowest point B

Em_f = K = ½ m v²

energy is conserved

Em₀ = Em_f

mg h = K

to find the height let's use trigonometry

at point A

cos 35 = x / L

x = L cos 35

so at the height is

h = L - L cos 35

h = L (1-cos 35)

we substitute

K = m g L (1 -cos 35)

we substitute in equation 1

W = m g L (1 -cos 35)

let's calculate

W = 0.500 9.8 0.950 (1 - cos 35)

W = 0.842 J

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3 years ago

How do the positions of the sun and moon affect what people do?

Verdich [7]

-- The position of the sun was originally the primary influence in determining
when people went to sleep and when they woke up. Although it no longer
directly influences us, that pattern is so deeply ingrained in our make-up
that our behavior still largely coincides with the positions of the sun.

-- The position of the Moon was originally the primary influence in determining
the cycle of human female physiology. Although it no longer directly influences
us, that pattern is so deeply ingrained in human make-up that the female cycle
still largely coincides with the positions of the Moon.

6 0

3 years ago

A constant force of 8N acting on an object displaces it through a distance of 3.0 m in the direction of force. Calculate work-do

sweet-ann [11.9K]

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{Work\:done=24\:J}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given :}}} \\ \tt: \implies Constant \: force(F) = 8 \: N \\ \\ \tt: \implies Displacement(s) = 3 \: m \\ \\ \red{\underline{\bold{To \: Find : }}} \\ \tt: \implies Work \: Done(W.D) = ?$

• <u>According to given question</u> :

$\green{ \star} \tt \: \theta \: = 0 \degree \: \: \: \: (Angle \: between \: force \: and \: displacement) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Work \: Done = FS \: cos \: \theta \\ \\ \tt: \implies Work \: Done = 8 \times 3 \times cos \:0 \degree \\ \\ \green{ \circ} \tt \: cos \: 0 \degree = 1 \\ \\ \tt: \implies Work \: Done =24 \times 1 \\ \\ \green{\tt: \implies Work \: Done =24 \: J}$

5 0

2 years ago

The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.6 square inches. If a force of 5.6 lb

balu736 [363]

Answer:

16.8 lb is the force on the brake pad of one wheel.

Explanation:

Force applied on the piston = $F_1=5.6 lb$

Area of the piston = $A_1=0.6 inches^2$

Force applied on the brakes = $F_2$

Area of the brakes = $A_2=1.8 inches^2$

Applying Pascal's law: 'For an incompressible fluid pressure at one surface is equal to the pressure at other surface'.

$\frac{F_1}{A_2}=\frac{F_2}{A_2}$

$F_2=\frac{5.6 lb\times 1.8 inhes^2}{0.6 inches^2}=16.8 lb$

16.8 lb is the force on the brake pad of one wheel.

5 0

3 years ago