Distance is speed x time. Half of the trip is 5.8/2 = 2.9hrs.
640 x 2.9 = 1856mi
580 x 2.9 = 1682mi
1856mi+1682mi=3538mi.

You could also calculate her average speed. This is easy since it was divided in two equal time slices. Average Speed = (640+580)/2 = 610mi/hr
Now 610mi/hr x 5.8hrs = 3538mi

7 0

3 years ago

A 48.0-kg astronaut is in space, far from any objects that would exert a significant gravitational force on him. He would like t

marusya05 [52]

Answer:

The astronaut is moving at a speed of 0.36m/s

Explanation:

Speed here corresponds to velocity

The astronaut's mass = 48kg

velocity of astronaut = ?

mass of socket = 0.72kg

velocity of socket = 5m/s

mass of the spanner = 0.8kg

velocity of spanner = 8m/s

change in time = 0.05 -0 = 0.05sec

mass of the mallet = 1.2kg

velocity of mallet = 6m/s

change in time = 9.9 -0 = 9.9sec

To find the astronaut velocity, we would calculate the total momentum which is the astronaut.

∑momentum (M) = ∑astronaut momentum

∑M = ∑astronaut M

∑astronaut M = M of socket + M of spanner + M of mallet

momentum = mass × velocity

(mass × velocity)of astronaut = (0.72×5) + (0.8×8) + (1.2×6)

48 × velocity of astronaut= 3.6 + 6.4 + 7.2

48 × velocity of astronaut= 17.2

velocity of astronaut = 17.2/48

velocity of astronaut = 0.36m/s

The astronaut is moving at a speed of 0.36m/s

5 0

3 years ago

The cars on an amusement-park ride travel at a constant velocity of 4.0 m/s on a circular track

lys-0071 [83]

V^2=u^2+2as
V=0
a =-u^2/2s
a=[4]^2/2[4]
a=-2m/s^2

7 0

3 years ago

Compare the time period of two simple pendulums of length 4m and 16m at a place.

Vlad1618 [11]

Answer:

the period of the 16 m pendulum is twice the period of the 4 m pendulum

Explanation:

Recall that the period (T) of a pendulum of length (L) is defined as:

$T=2\,\pi\,\sqrt{ \frac{L}{g} }$

where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

$T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2$

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.

5 0

3 years ago