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3 years ago

How far does a car travel in 0.75h if it is moving at a constant speed of 88km/h

2 answers:

Citrus2011 [14]3 years ago

6 0

V=D/T - formula for speed
88 km/h = D / 0.75 hours - multiply speed and time
66 km= D
D is distance
D= 66 km
Hope this helps

Svetlanka [38]3 years ago

3 0

Answer:

A car will travel 66 km in 0.75 hours with 88 km/hour speed.

Explanation:

$Speed=\frac{\text{Distance covered by body}}{\text{time taken by the body}}$

$S=\frac{D}{T}$

We have:

Distance covered by the car = D = ?

Time taken by the car to cover D distance = T= 0.75 hours

Speed of the car = S = 88 km/h

$88 km/h=\frac{D}{0.75 h}$

$D=88 km/h\times 0.75 h=66 km$

A car will travel 66 km in 0.75 hours with 88 km/hour speed.

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For this discussion, you will work in groups to answer the questions. In a video game, airplanes move from left to right along t

Mariulka [41]

Answer:

When fired from (1,3) the rocket will hit the target at (4,0)

When fired from (2, 2.5) the rocket will hit the target at (12,0)

When fired from (2.5, 2.4) the rocket will hit the target at $(\frac{35}{2},0)$

When fired from (4,2.25) the rocket will hit the target at (40,0)

Explanation:

All of the parts of the problem are solved in the same way, so let's start with the first point (1,3).

Let's assume that the rocket's trajectory will be a straight line, so what we need to do here is to find the equation of the line tangent to the trajectory of the airplane and then find the x-intercept of such a line.

In order to find the line tangent to the graph of the trajectory of the airplane, we need to start by finding the derivative of such a function:

$y=2+\frac{1}{x}$

$y=2+x^{-1}$

$y'=-x^{-2}$

$y'=-\frac{1}{x^{2}}$

so, we can substitute the x-value of the given point into the derivative, in this case x=1, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(1)^{2}}$

m=y'=-1

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-3=-1(x-1})$

$y-3=-1x+1$

$y=-x+1+3$

$y=-x+4$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-x+4=0$

and solve for x

x=4

so, when fired from (1,3) the rocket will hit the target at (4,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2, 2.5)

so, we can substitute the x-value of the given point into the derivative, in this case x=2, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(2)^{2}}$

$m=y'=-\frac{1}{4}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.5=-\frac{1}{4}(x-2})$

$y-2.5=-\frac{1}{4}x+\frac{1}{2}$

$y=-\frac{1}{4}x+\frac{1}{2}+\frac{5}{2}$

$y=-\frac{1}{4}x+3$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{1}{4}x+3=0$

and solve for x

x=12

so, when fired from (2, 2.5) the rocket will hit the target at (12,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2.5, 2.4)

so, we can substitute the x-value of the given point into the derivative, in this case x=2.5, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(2.5)^{2}}$

$m=y'=-\frac{4}{25}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.4=-\frac{4}{25}(x-2.5})$

$y-2.4=-\frac{4}{25}x+\frac{2}{5}$

$y=-\frac{4}{25}x+\frac{2}{5}+2.4$

$y=-\frac{4}{25}x+\frac{14}{5}$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{4}{25}x+\frac{14}{5}=0$

and solve for x

$x=\frac{35}{20}$

so, when fired from (2.5, 2.4) the rocket will hit the target at $(\frac{35}{2},0)$

Now, let's calculate the coordinates where the rocket will hit the target if fired from (4, 2.25)

so, we can substitute the x-value of the given point into the derivative, in this case x=4, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(4)^{2}}$

$m=y'=-\frac{1}{16}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.25=-\frac{1}{16}(x-4})$

$y-2.25=-\frac{1}{16}x+\frac{1}{4}$

$y=-\frac{1}{16}x+\frac{1}{4}+2.25$

$y=-\frac{1}{16}x+\frac{5}{2}$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{1}{16}x+\frac{5}{2}=0$

and solve for x

$x=40$

so, when fired from (4,2.25) the rocket will hit the target at (40,0)

I uploaded a graph that represents each case.

8 0

3 years ago

These the flow of electrons (the current) and where some of the electrons' energy gets converted into heat. (Lessons 5.01-5.03)

liq [111]

Answer:

Conductors.

Explanation:

Conductors allow the flow of electrons i. e. the current and where some of the electrons energy gets converted into heat while some electrons energy converted into light. Conductors are the materials which has the ability to allow heat and electricity to flow through them. Metals such as gold, copper, Silver, Aluminum, Mercury, Steel, Iron and salty water etc are good conductors of heat and electricity.

3 0

3 years ago

Explanation A 5000 kg rocket is at rest in deep space. The rocket burns fuel pushing 10kg of exhaust gases rearward at 4000 m/s.

katovenus [111]

Answer:

F = 4000 N

Explanation:

given,

mass of rocket (M)= 5000 Kg

10 Kg gas burns at speed (m)= 4000 m/s

time = 10 s

average force = ?

at the end the rocket is at rest

by conservation of momentum

M v + m v' = 0

5000 x v - 10 x 4000 = 0

5000 v = 40000

v = 8 m/s

speed of rocket = 8 m/s

now,

we know

change in momentum = F x Δ t

$F = \dfrac{m(v_i-v_f)}{\Delta t}$

$F = \dfrac{5000(8-0)}{10}$

F = 4000 N

Hence, the average force applied to the rocket is equal to F = 4000 N

4 0

4 years ago

Read 2 more answers

Without an atmosphere, the equatorial curve would show minimum daily values on the solstices in June when the sub-solar point is

lapo4ka [179]

Without an atmosphere, the equatorial curve would show minimum daily values on the solstices in June when the sub-solar point is located at 23.5°N and in December when the sub-solar point is at 23.5°S latitude.

Explanation:

At the sub-solar point, the sun strikes directly at the surface with an angle of 90 degrees at a given point.

Solistice refers to that point in time when the sun’s zenith is located at the farthest point from the equator.

During summer solistice on June 21, the sun’s zenith reaches northernmost point, sub-solar point is fixed at 23.5°S Tropic of Cancer making the earth tilt 23.4 degrees

During winter soliscitse on December 21, the sub-solar point is fixed at) Tropic of Capricorn.

3 0

3 years ago

What velocity must a car with a mass of 1280 kg have in order to have the same momentum as a 2230?

sveta [45]

Momentum = (mv).
<span>(2110 x 24) = 50,640kg/m/sec. truck momentum. </span>
<span>Velocity required for car of 1330kg to equal = (50,640/1330), = 38m/sec</span>

8 0

3 years ago