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Igoryamba
4 years ago
9

You and a friend each drive 58 km. You travel at 89. km/h, your friend at 94 km/h. How long will your friend wait for you at the

end of the trip?
Physics
1 answer:
alexgriva [62]4 years ago
5 0

Answer:

2.1 minutes/ 126 seconds

Explanation:

Distance = speed x time

We can rearrange this equation for time:

Time = Distance/Speed

<u>For you (89km/h):</u>

Distance = 58km, Speed = 89km/h

Therefore time = 58/89 = 0.652 (3dp) hours

<u>For friend:</u>

Distance = 58km, Speed = 94km/h

Therefore time = 58/94 = 0.617 (3dp) hours

Difference in time = 0.652 - 0.617 = 0.035 hours.

Convert to minutes: 0.035 x 60 (because 60 min in hour) = 2.1

Your friend will be waiting for 2.1 minutes, or 126 seconds (2.1 x 60).

Hope this helped!

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Dimension of force of gravity is​
statuscvo [17]

Answer:

M^{-1}L^{3}T^{-2}

Explanation:

Force has the unit N for Newton.

A Newton is m*\frac{kg}{s^{2} } , mass*kilograms/seconds^2.

Dimensions in Physics include things such as Mass, Length, and Time.

kg is Mass

m is Length

s^2 is Time

So, since a Newton has all of these, it has T, L, and M.<em> </em><em>Force has dimensions  </em>T^{-2}LM<em />

However, you must now consider gravitational force.

Law of Gravitation states:

F=\frac{Gm_{1}m_{2}r  }{r^{2} }

Convert that into a fraction with dimensions:

F=\frac{GM^{2} L}{L^{2} }

Remember force is  T^{-2}LM

Use the dimensions of force to find gravitational force, or G:

F=\frac{GM^{2} L}{L^{2} }

T^{-2}LM=\frac{GM^{2} L}{L^{3} }

G=\frac{(T^{-2}LM)(L^{2}) }{M*M}

G=\frac{(T^{-2}L^{3} M) }{M^{2} }

G=M^{-1}L^{3}T^{-2}

5 0
3 years ago
A fountain can hold 53,000 deciliters of water. how many kiloliters is this ?
zubka84 [21]
5.3 kiloliters is how much water a fountain can hold.
3 0
3 years ago
The density of mercury is 13.6 g/cm3 calculate the mass of 1 cm3 of mercury
7nadin3 [17]

the mass of 1cm3 of mercury is 13.6g because

the formula of density is mass/volume and when paste the number in it. you get the answer

5 0
3 years ago
Using a radar gun, you emit radar waves at a frequency of 6.2 GHz that bounce off of a moving tennis ball and recombine with the
Keith_Richards [23]

Answer:

23.4 m/s

Explanation:

f = actual frequency of the wave = 6.2 x 10⁹ Hz

f_{app} = frequency observed as the ball approach the radar

f_{rec} = frequency observed as the ball recede away from the radar

V = speed of light

v = speed of ball

B = beat frequency = 969 Hz

frequency observed as the ball approach the radar is given as

f_{app}=\frac{f(V+v)}{V}                                 eq-1

frequency observed as the ball recede the radar is given as

f_{rec}=\frac{f(V-v)}{V}                                  eq-2

Beat frequency is given as

B = f_{app} - f_{rec}

Using eq-2 and eq-1

B = \frac{f(V+v)}{V}- \frac{f(V-v)}{V}

inserting the values

969 = \frac{(6.2\times 10^{9})((3\times 10^{8})+v)}{(3\times 10^{8})}- \frac{(6.2\times 10^{9})((3\times 10^{8})-v)}{(3\times 10^{8})}

v = 23.4 m/s

8 0
3 years ago
A slingshot is fired with an initial velocity of 100 m/s at an angle of 55° follows a parabolic trajectory and hits a stationary
erma4kov [3.2K]

Answer:

The horizontal speed with which the slingshot hits the balloon is approximately 57.358 m/s

Explanation:

The given parameters of the slingshot motion are;

The initial velocity of the slingshot, v = 100 m/s

The angle (to the horizontal) at which the slingshot is fired makes, θ = 55°

The path of the slingshot which hits the stationary balloon at the top of its flight = Parabolic trajectory

The horizontal component of the velocity = vₓ = v·cos(θ) = Constant

vₓ = 100 × cos(55°) ≈ 57.358

The horizontal speed with which the slingshot hits the balloon = vₓ ≈ 57.358 m/s.

4 0
3 years ago
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