All categories

3 years ago

How much power is used by a car engine if it does 48496 J of work in 4 min?

Physics

2 answers:

elena-14-01-66 [18.8K]3 years ago

8 0

Hope this will help u.....✌✌✌

cupoosta [38]3 years ago

6 0

Answer:

hope it helps uh.............

You might be interested in

A rope is tied to a box and used to pull the box 2.3 m along a horizontal floor. The rope makes an angle of 30∘ with the horizon

Savatey [412]

Answer:

Answered

Explanation:

A) The work done by gravity is zero because displacement and the gravitational force are perpendicular to each other.

W= FS cosθ

θ= 90 ⇒cos90 = 0 ⇒W= 0

B) work done by tension

W= Tcosθ×S= 5cos30×2.30= 10J

C) Work done by friction force

W= f×s=1×2.30= 2.30 J

D) Work done by normal force is Zero because the displacement and the normal force are perpendicular to each other.

E) The net work done= Work done by tension in the rope - frictional work

=10-2.30= 7.7 J

6 0

2 years ago

In the diagnostic radiology energy range (which includes mammography) from 23 to 150 kVp, which of the following tissues possess

Semmy [17]

Answer:

Bone

Explanation:

Diagnostic radiology include the use of non-invasive imaging scans to diagnose a patient.

The voltages used in diagnostic tubes range from roughly 20 kV to 150 kV and thus the highest energies of the X-ray photons range from roughly 20 keV to 150 keV.

The tests and equipment used sometimes involves low doses of radiation to create highly detailed images of an area.

4 0

2 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s$

$v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s$

Time taken to reach 1180 m is 11.29 seconds

$v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s$

Time the rocket will keep going up after the engines shut off is 13.06 seconds.

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m$

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s$