25 POINTS please help!

Tourists covered 255 km for a 4-hour ride by car and a 7-hour ride by train. what is the speed of the train, if it is 5 km/h gre

LekaFEV [45]

The distance covered by car is equal to (assuming it is moving by uniform motion) the product between the car's speed and the time of the car ride, 4 h:
$S_c = v_c t_c$
where
$v_c$ is the car's speed
$t_c = 4 h$ is the duration of the car ride

Similarly, the distance covered by train is equal to the product between the train's speed and the duration of the train ride, 7 h:
$S_t = v_t t_t$

The total distance covered is S=255 km, which is the sum of the distances covered by car and train:
$S=255 km = S_c + S_t$
which becomes
$255 = 4 v_c + 7 v_t$ (1)
we also know that the train speed is 5 km/h greater than the car's speed:
$v_t = 5 + v_c$ (2)

If we put (2) into (1), we find
$255 = 4v_c + 7(5+v_c)$
and if we solve it, we find
$v_c = 20 km/h$
$v_t = 25 km/h$

So, the car speed is 20 km/h and the train speed is 25 km/h.

4 0

3 years ago

EASY BRAINLIEST PLEASE HELP!!

Rudiy27

Answer:

I think the awnser is B (but don't qoute me on that) if its right then yay but if its wrong im sorry

Explanation:

5 0

3 years ago

Read 2 more answers

Can anyone explain how tides work

Mila [183]

The position of the sun and the moon affect how high the tide is

6 0

3 years ago

Read 2 more answers

James threw a ball vertically upward with a velocity of 41.67ms-1 and after 2 second David threw a ball vertically upward with a

Reptile [31]

Answer:

When have passed 3.9[s], since James threw the ball.

Explanation:

First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.

We'll use the kinematics equations to find these two unknowns.

$y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can find the velocity after 2 seconds.

$v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

$y=y_{o} +v_{o} *t-\frac{1}{2} *g*t^{2} \\y=63.72 +22.05 *t-\frac{1}{2} *(9.81)*t^{2} \\\\y=63.72 +22.05 *t-4.905*t^{2} \\$

As we can see the equation is based on Time (t).

Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t

$y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]$

Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.

Time = 2 + 1.9 = 3.9[s]

4 0

3 years ago

A proposed communication satellite would revolve around the earth in a circular orbit in the equatorial plane at a height of 358

aleksandr82 [10.1K]

Answer:

Period is 86811.5 seconds.

Explanation:

${ \boxed{ \bf{T {}^{2} = (\frac{4 {\pi}^{2} }{GM}) {r}^{3} }}}$

${ \tt{T {}^{2} = \frac{4 {(3.14)}^{2} }{(6.6 \times {10}^{ - 11} ) \times (5.98 \times {10}^{24} )} \times {((35880\times {10}^{3}) } + (6370 \times {10}^{3} )) {}^{3} }} \\ \\ { \tt{T {}^{2} = 7.54 \times {10}^{9} }} \\ { \tt{T = \sqrt{7.54 \times {10}^{9} } }} \\ { \tt{T = 86811.5 \: seconds}}$

6 0

3 years ago