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3 years ago

You drive your car in a straight line at 15 m/s for 10 kilometers, then at 25 m/s for another 10 kilometers.

Physics

1 answer:

Vikki [24]3 years ago

5 0

Answer:

A) Average speed = 18.75 m/s

B) More time is spent at 15 m/s than at 25 m/s.

Explanation:

Let the first distance be d1 and the second distance be d2.

We are given;

d1 = 10 km = 10000 m

d2 = 10 km = 10000 m

Speed; v1 = 15 m/s

Speed; v2 = 25 m/s

Now, the formula for distance is; Distance = speed x time

Thus:

d1 = v1 x t1

t1 = d1/v1 = 10000/15 = 666.67 seconds

Also,

d2 = v2 x t2

t2 = d2/v2 = 10000/25 = 400 seconds

Average speed = total distance/total time = (10000 + 10000)/(666.67 + 400) = 18.75 m/s

From earlier, since t1 = 666.67 seconds and t2 = 400 seconds, then;

More time at 15 m/s than at 25 m/s.

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Physics, calculating net force. Please work it out for me

Margarita [4]

Answer:

F = 2985.125 N

Explanation:

Given that,

The radius of curvature of the roller coaster, r = 8 m

Speed of Micheal, v = 17 m/s

Mass of body, m = 65 kg We need to find the net force acting on Micheal. Net force act the bottom of the circle is given by :

$F=\dfrac{mv^2}{r}+mg\\\\F=m(\dfrac{v^2}{r}+g)\\\\=65(\dfrac{17^2}{8}+9.8)\\\\=2985.125\ N$

So, the net force is 2985.125 N.

3 0

3 years ago

Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated

Artemon [7]

Answer: Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN

Explanation: To find the answer we need to know more about the Newton's law of gravitation.

<h3>What is Newton's law of gravitation?</h3>

Gravitation is the force of attraction between any two bodies.
Every body in the universe attracts every other body with a force.
This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between these two masses.
Mathematically we can expressed it as,

$F=\frac{GMm}{r^2} \\where, G=6.67*10^-11Nm^2kg^-2$

<h3>How to solve the problem?</h3>

Here, we have given with the data's,

$M=8.22*10^9kg\\m=1.38*10^8 kg\\r=1.43*10^3m$

Thus, the force of attraction between these two bodies will be,

$F=6.67*10^-11*\frac{8.22*10^9*1.38*10^8}{1.43*10^3} =52.9kN$

Thus, if two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg and, If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN.

Learn more about the Newton's law of gravitation here:

brainly.com/question/28045318

#SPJ4

6 0

1 year ago

give an example of motion in which displacement of the particle is zero but acceleration is not zero in journey?

BartSMP [9]

Motion of a ball thrown by a person upwards and caught after some time is an example of motion in which displacement of the particle is zero but acceleration is not zero in journey.

The displacement of the ball is zero because the starting and end point of the motion are same, i.e, the person's hands.During its motion, the acceleration of ball is constant and non zero called as acceleration due to gravity, g= -9.8 m/s². The velocity of ball is continuously changing. It first decreases during the upward motion of the ball and then increases during the downward journey.The acceleration remains constant and non zero all the time.

4 0

2 years ago

A parallel-plate capacitor has 2.10 cm × 2.10 cm electrodes with surface charge densities ±1.00×10-6 C/m2. A proton traveling pa

Darya [45]

Answer:

x=0.53 $x10^{-3} m$

Explanation:

Using Gauss law the field is uniform so

E=ζ/ε

Charge densities ⇒ζ=1. $x10x^{-6} \frac{C}{m^{2}}$

ε=8.85 $x10^{-12} \frac{C^{2}}{n*m^{2}}$

$E=\frac{1x10^{-6}\frac{C}{m^{2}}}{8.85x^{-12}\frac{C^{2} }{N*m^{2}}} \\E=0.11299 x10^{-6} \frac{N}{C}$

Force of charge is

$F_{q}=q*E\\F_{q}=1.6x10^{-19}C*0.11299x10^{6}\frac{N}{C} \\F_{q}=1.807x10^{-14} N$

$F_{q}=m*a\\a=\frac{F_{q}}{m}=\frac{1.807x10^{-13}N}{1.67x10^{-27}}\\ a=1.082x0^{14} \frac{m}{s^{2}} \\t=\frac{x}{v}\\ x=2.1cm\frac{1m}{100cm}=0.021m \\v=6.7x10^{6}\frac{m}{s} \\ t=\frac{0.021m}{6.7x10^{6}\frac{m}{s}} \\t=3.13x10^{-9}s$

So finally knowing the acceleration and the time the distance can be find using equation of uniform motion

$x_{f}=x_{o}+\frac{1}{2}*a*t^{2}\\ x_{o}=0\\x_{f}=\frac{1}{2} a*t^{2}=\frac{1}{2}*1.082x10^{14}\frac{m}{s^{2} } *(3.134x^{-9}s)^{2} \\x_{f}=0.53x^{-3}m$

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2 years ago

The (blank) of vibration of a wave is defined as that which has the lowest frequency

saveliy_v [14]

Fundamental frequency

6 0

3 years ago

Read 2 more answers