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Naddik [55]
3 years ago
6

A body travels 10 meters during the first 5 seconds of its travel, and it travels a total of 30 meters over the first 10 seconds

of its travel. What is
its average speed during the time from t = 5 seconds to t = 10 seconds?
ОА.
2 meters/second
B.
3 meters/second
C.
4 meters/second
D.
5 meters/second
E.
6 meters/second
Physics
1 answer:
Inessa05 [86]3 years ago
5 0

Answer:

A body travels 10 meters during the first 5 seconds of its travel,and a total of 30 meters over the first 10 seconds of its travel

20miles / 5sec = 4miles /sec would be the average speed for the last 20 m

Explanation:

The answer is 4 m/s.

In the first 5 seconds, a body travelled 10 meters. In the first 10 seconds of the travel, the body travelled a total of 30 meters, which means that in the last 5 seconds, it travelled 20 meters (30m + 10m).

The relation of speed (v), distance (d), and time (t) can be expressed as:

v = d/t

We need to calculate the speed of the second 5 seconds of the travel:

d = 20 m (total 30 meters - first 10 meters)

t = 5 s (time from t = 5 seconds to t = 10 seconds)

Thus:

v = 20m / 5s = 4 m/s

PLEASE GIVE BRAINIEST!! HOPE THIS HELPS

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A person jogs for 4.0 km in 32 mins,then 2.0km in 22mins,and finallu 1.0 km in 16 mins.whqt is the joggers average speed in km p
blondinia [14]

Average speed = (total distance covered) / (time to cover the distance)

total distance covered = (4km + 2km + 1km) = 7 km

time to cover the distance = (32min + 22min + 16min) = 70 min

Average speed = (7 km) / (70 min)

Average speed = 0.1 km/minute

5 0
3 years ago
The scientist to first introduce the concept of inertia was
skelet666 [1.2K]
The first scientist to introduce the concept of inertia was Galileo
4 0
3 years ago
Which shows a decrease in fluid pressure? A. A fan is turned from high speed to low speed. B. Oxygen is compressed as it is put
Volgvan

Answer:

Option A.

A fan is turned from high speed to low speed.

Explanation:

It is important to note that air is also a fluid.

In a system, static pressure of air increases with the speed of rotation of the fan. This is because when the speed of the fan is increased, the force with which it is pushing the air molecules is increased. Since pressure is a relationship between force and area, the pressure of the air molecules will be increased.

Conversely, when the speed of the fan is reduced, the priming force on the air molecules will be reduced, hence the pressure of the air will drop.

This makes option A the correct option

8 0
3 years ago
Read 2 more answers
it takes 90 j of work to stretch a spring 0.2 m from its equilibrium position. How muc work is needed to stretch it an additiona
Vinvika [58]

Work needed: 720 J

Explanation:

The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the stretching of the spring from the equilibrium position

In this problem, we have

E = 90 J (work done to stretch the spring)

x = 0.2 m (stretching)

Therefore, the spring constant is

k=\frac{2E}{x^2}=\frac{2(90)}{(0.2)^2}=4500 N/m

Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of

x = 0.2 + 0.4 = 0.6 m

Substituting,

E'=\frac{1}{2}kx^2=\frac{1}{2}(4500)(0.6)^2=810 J

Therefore, the additional work needed is

\Delta E=E'-E=810-90=720 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

7 0
2 years ago
How much work must be done to stop a 1100-kg car traveling at 112 km/h?(Hint: You will need to convert the speed first.)Answer:
zimovet [89]

According to the Work-Energy Theorem, the work done on an object is equal to the change in the kinetic energy of the object:

W=\Delta K

Since the car ends with a kinetic energy of 0J (because it stops), then the work needed to stop the car is equal to the initial kinetic energy of the car:

K=\frac{1}{2}mv^2

Replace m=1100kg and v=112km/h. Write the speed in m/s. Remember that 1m/s = 3.6km/h:

\begin{gathered} K=\frac{1}{2}(1100kg)\left(112\frac{km}{h}\times\frac{1\frac{m}{s}}{3.6\frac{km}{h}}\right)^2=532,345.679...J \\  \\ \therefore K\approx532,346J \end{gathered}

Therefore, the answer is: 532,346 J.

5 0
10 months ago
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