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3 years ago

14

A 15N block is sliding at constant speed across the frictionless floor of an ice rink.

1 answer:

Gnesinka [82]3 years ago

8 0

Answer:

so, why we have to find here..

<h3>stay safe healthy and happy.</h3>

You might be interested in

Only the healthy foods we eat provide<br> energy to our body<br> true<br> false

Ilia_Sergeevich [38]

Answer:

True

Explanation:

Yes absolutely the food which provede us energy is termed as healthy..

7 0

3 years ago

Engineers at the Space Centre must determine the net force needed for a rockets engine to achieve an acceleration of 70 m/s2. As

Troyanec [42]

Answer:

3,150,000N

Explanation:

According to Newton's second law;

F = mass * acceleration

Given

Mass = 45000kg

acceleration = 70m/s^2

Substitute

F = 45000 * 70

F = 3,150,000N

Hence the force required to be produced by the rocket engines is 3,150,000N

8 0

3 years ago

A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli

Art [367]

Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

a) satellite's orbital velocity

centripetal force acting on satellite = $\dfrac{mv^2}{r}$

gravitational force = $\dfrac{GMm}{r^2}$

equating both the above equation

$\dfrac{mv^2}{r} = \dfrac{GMm}{r^2}$

$v = \sqrt{\dfrac{GM}{r}}$

$v = \sqrt{\dfrac{6.67 \times 10^{-11}\times 5.972 \times 10^{24}}{2 \times 6.4 \times 10^6}}$

v = 5578.5 m/s

b) $T= \dfrac{2\pi\ r}{v}$

$T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}$

$T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}$

T = 14416.92 s

$T = \dfrac{14416.92}{3600}\ hr$

T = 4 hr

c) gravitational force acting

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

$F = \dfrac{6.67 \times 10^{-11}\times 545 \times 5.972 \times 10^{24} }{(6.46 \times 10^6)^2}$

F = 5202 N

4 0

3 years ago

If a ball that is 10 meters above the ground is thrown horizontally at 5.51 meters per second. a. how long will it take for the

GalinKa [24]

Answer:

a. t = 1.43 s

b. d = 7.88 m

Explanation:

a. The time of flight can be found using the following equation:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final height = -10 m

$y_{0}$ : is the initial height = 0

$v_{0_{y}}$ : is the initial speed in the vertical direction = 0

g: is the acceleration due to gravity = 9.81 m/s²

By solving the above equation for "t" we have:

$t = \sqrt{\frac{2y_{f}}{g}} = \sqrt{\frac{2*10 m}{9.81 m/s^{2}}} = 1.43 s$

Hence, the ball will hit the ground in 1.43 s.

b. The distance in the horizontal direction can be found as follows:

$x_{f} = x_{0} + v_{0}t + \frac{1}{2}at^{2}$

Where:

x₀: is the initial position in the horizontal direction = 0

a: is the acceleration in the horizontal direction = 0 (it is moving at constant speed)

$x_{f} = 5.51 m/s*1.43 s = 7.88 m$

Therefore, the ball will travel 7.88 m before it hits the ground.

I hope it helps you!

4 0

3 years ago

Can someone solve this problem and explain to me how you got it

evablogger [386]

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒ $F\alpha\frac{q1*q2}{r^{2}}$

∴ $F=k\frac{q1*q2}{r^{2}}$

where $F$ is the force of attraction or repulsion

$k$ is Coulumb's constant= $9*10^{9}Nm^{2}C^{-2}$

$q1$ and $q2$ are the magnitude of the charges

$r$ is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

$F=9*10^{9}*\frac{0.041*0.029}{12^{2}}$

$F=74312.5N$

Question6:

Given: q1= $3*10^{-18}C$ , r=5 m, F= $4*10^{-11}N$

therefore by Coulumb's law,

$4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}$

⇒ $q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C$

4 0

3 years ago