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3 years ago

If a car increases its velocity from zero to 60 m/s in 10 seconds, its acceleration is

1 answer:

5 0

We know that a=vf_vi/t equals equation "a" . Where a is the acceleration of the body , vf is the final velocity , vi is the initial velocity and t is equal to time . Since vi equals o m/s , vf equals to 60 m/s and t equals 10 s. Put in equation "a". a=60-0/10 =6m/s2

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Select the choice that best completes the following sentence. Simple machines A. reduce the amount of energy needed to do work.

boyakko [2]

It is actually are tools that make work easier C because i just had it on study island

7 0

3 years ago

Read 2 more answers

How much force does an 88kg astronaut exert on his chair while accelerating straight up at 10 m/s^2

Fed [463]

Force=mass*acceleration. So 88kg*10 m/s^2=880 newtons

6 0

3 years ago

New 5G networks utilize millimeter-wave radiation. Millimeter-wave radiation refers to electromagnetic waves with frequencies in

seraphim [82]

Answer:

It corresponds to 1mm-10 mm range.

Explanation:

Electromagnetic waves (such as the millimeter-wave radiation) travel at the speed of light, which is 3*10⁸ m/s in free space.
As in any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:

$v = \lambda * f (1)$

Replacing v= c=3*10⁸ m/s, and the extreme values of f (which are givens), in (1) and solving for λ, we can get the free-space wavelengths that correspond to the 30-300 GHz range, as follows:

$\lambda_{low} = \frac{c}{f_{high}} = \frac{3e8m/s}{300e9Hz} = 1 mm (2)$

$\lambda_{high} = \frac{c}{f_{low}} = \frac{3e8m/s}{30e9Hz} = 10 mm (3)$

4 0

3 years ago

At noon, ship A is 140 km west of ship B. Ship A is sailing east at 30 km/h and ship B is sailing north at 25 km/h. How fast (in

Luba_88 [7]

Answer:

The rate of change of distance between the two ships is 18.63 km/h

Explanation:

Given;

distance between the two ships, d = 140 km

speed of ship A = 30 km/h

speed of ship B = 25 km/h

between noon (12 pm) to 4 pm = 4 hours

The displacement of ship A at 4pm = 140 km - (30 km/h x 4h) =

140 km - 120 km = 20 km

(the subtraction is because A is moving away from the initial position and the distance between the two ships is decreasing)

The displacement of ship B at 4pm = 25 km/h x 4h = 100 km

Using Pythagoras theorem, the resultant displacement of the two ships at 4pm is calculated as;

r² = a² + b²

r² = 20² + 100²

r = √10,400

r = 101.98 km

The rate of change of this distance is calculated as;

r² = a² + b²

r = 101.98 km, a = 20 km, b = 100 km

$2r(\frac{dr}{dt} ) = 2a(\frac{da}{dt} ) + 2b(\frac{db}{dt} )\\\\r(\frac{dr}{dt} ) = a(\frac{da}{dt} ) + b(\frac{db}{dt} )\\\\101.98(\frac{dr}{dt} ) = 20(-30 ) + 100(25 )\\\\101.98(\frac{dr}{dt} ) = -600 + 2,500\\\\101.98(\frac{dr}{dt} ) = 1900\\\\\frac{dr}{dt} = \frac{1900}{101.98} = 18.63 \ km/h$

5 0

3 years ago

Cal ulate a moment of force o. 50 meter distance and 10 newton force

lina2011 [118]

Answer:

500N/M

Explanation:

given that

force=10N

distance=50M

moment=force*distance

=10×50=500j

8 0

3 years ago