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MA_775_DIABLO [31]
3 years ago
10

Titus drives his jetski a distance of 1000 meters in 7.045 seconds. How fast was he moving in meters per second? How fast was he

moving in miles per hour?
Physics
1 answer:
Ksenya-84 [330]3 years ago
5 0

Answer:

a) v = 141.9 m/s

b) v = 317.4 miles/h

Explanation:

a) How fast was he moving in meters per second?

v = \frac{d}{t} = \frac{1000 m}{7.045 s} = 141.9 m/s

Hence, the jet ski is moving at 141.9 meters per second.

b) How fast was he moving in miles per hour?

v = \frac{d}{t} = \frac{1000 m}{7.045 s} = 141.9 m/s*\frac{3600 s}{1 h}*\frac{1 mile}{1609.34 m} = 317.4 miles/h      

Therefore, the jet ski is moving at 317.4 miles per hour.

I hope it helps you!

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Silicon (chemical symbol Si) is located in Group 14, Period 3. Which is silicon
VARVARA [1.3K]

Answer:

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2 years ago
What is your wheel and axle
Crank

Answer:

Pls mark me brainliest

Explanation:

The wheel and axle is a simple machine consisting of a wheel attached to a smaller axle so that these two parts rotate together in which a force is transferred from one to the other.

8 0
2 years ago
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Sallys physical education teacher timed her run and recorded the time and distance in the table below. What is her average speed
Fudgin [204]
Answer: B) 2.5 m/s

Explanation: Find the average of the time and distance, and see how far they go in only 1 second.


1 + 2 + 3 + 4 + 5 = 15
15 divided by 5 = 3

3 seconds

2 + 5 + 7 + 10 + 12 = 36
36 divided by 5 = 7.2


7.2m per 3 seconds.

7.2 divided by 3 = 2.4

Therefore, the answer is technically 2.4m/s
4 0
3 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
3 years ago
Why is it important to understand forces?
yanalaym [24]
If people never learned forces, there would be a major gap in the world and how it works, let alone in physics...
as much as you don't wanna admit it, force is everywhere and you see it if not use it EVERY day in your life, something as simple as driving a car down the street or too school, your using force of your wheels to move your car, which is moving you
3 0
3 years ago
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