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3 years ago

HELP ASP PLEASED AND THANKS NO LINKS PLEASED AND NO FILES

Physics

1 answer:

SOVA2 [1]3 years ago

6 0

Answer:

just search up a ven-diagram and then try to draw it or trace it then use it for ur question

Explanation:

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Which refers to the rate of change in velocity?

alexira [117]

The answer is obvious acceleration lol sorry if it sounds mean

7 0

3 years ago

If a Ferrari, with an initial velocity of 10 m/s, accelerates at a rate of 50 m/s 2 for 3 seconds, what will its final velocity

Evgen [1.6K]

Answer:

160m/s

Explanation:

The Ferrari is moving by uniformly accelerated motion, with constant acceleration of a = 50 m/s^2, and initial velocity u = 10 m/s. The velocity at time t of the car is given by

v(t)=u+at

where

u = 10 m/s

a = 50 m/s^2

If we substitute t = 3 s into the equation, we can find the velocity of the car after 3 seconds:

v(3s) = 10m/s + (50m/s^2)(3s)=160m/s

3 0

3 years ago

Read 2 more answers

How many meters in 2.50 miles? (Use these two conversions: 1000 m = 1 km and 1.00 km = .621 mi )

Artyom0805 [142]

2.50 miles is equal to 4026 m.

<u>Explanation:</u>

As it is known that 1000 m =1 km and 1 km = 0.621 miles. So first we have to convert miles to km and then to metre as follows.

As 1 km = 0.621 miles, then

$\text { 1 miles }=\frac{1}{0.621} \mathrm{km}$

So, 2.50 miles will be equal to

$2.50 \text { miles }=\frac{2.50}{0.621} \mathrm{km}=4.026 \mathrm{km}$

Then, in order to get the answer in meters, we have to convert this km to meter by the conversion of 1000 m =1 km. So,

$1 \mathrm{km}=1000 \mathrm{m}$

Thereby,

$4.026 \mathrm{km}=4026 \mathrm{m}$

7 0

3 years ago

How many significant digits are in the number 204.0920<br>

monitta

Answer:

Seven

Explanation:

The rules for significant digits are:

Non-zero digits are always significant.
Zeros between significant digits are also significant.
Trailing zeros are significant only after a decimal point.

Here, the 2, 4, 9, and 2 are significant because they are non-zero digits.

The first two 0s are significant because they are between significant digits.

The last 0 is significant because it is a trailing zero after a decimal point.

Therefore, all seven digits are significant.

3 0

3 years ago

Read 2 more answers

Determine the gravitational potential energy, in kJ, of 3 m3 of liquid water at an elevation of 40 m above the surface of Earth.

melomori [17]

Explanation:

We will calculate the gravitational potential energy as follows.

$P.E_{1} = mgz_{1}$

$P.E_{1} = (\rho V)gz_{1}$

= $1000 kg/m^{3} \times 3 m^{3} \times 9.7 \times 40 m$

= 1164000 J

or, = 1164 kJ (as 1 kJ = 1000 J)

Now, we will calculate the change in potential energy as follows.

$\Delta P.E = mg(z_{2} - z_{1})$

= $\rho \times V \times g (z_{2} - z_{1})$

= $1000 \times 3 \times 9.7 (10 - 40)m$

= -873000 J

or, = -873 kJ

Thus, we can conclude that change in gravitational potential energy is -873 kJ.

4 0

3 years ago