3 years ago

PLZ hurry! Thank you

Physics

1 answer:

Alenkinab [10]3 years ago

3 0

Answer:do it again

Explanation:

It will work

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An object is dropped at a height of 6 m from the ground. How fast is it moving just before it hits the ground?

maw [93]

Explanation:

Given:

Δy = 6 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (9.8 m/s²) (6 m)

v ≈ 10.8 m/s

Round as needed.

3 0

3 years ago

A toy car of mass 8 kg initially moves with a speed of 5 m/s. How much work must be done on the car to increase its speed to 10

sukhopar [10]

ANSWER:

300 J

STEP-BY-STEP EXPLANATION:

To know the work required, we must calculate the work in both cases, the difference would be the amount of work necessary for the speed to increase. The work done is the same as the amount of energy increase. The formula for kinetic energy is:

$\frac{1}{2}m\cdot v^2$

We calculate in each case:

$\begin{gathered} \frac{1}{2}\cdot8\cdot10^2=400\text{ J} \\ \frac{1}{2}\cdot8\cdot5^2=100\text{ J} \end{gathered}$

We calculate the difference between the two to find out the work required:

$400-100=300\text{ J}$

The work required is 300 J

3 0

2 years ago

An aircraft is moving horizontally with a speed of 50m/s, at the height of 2km, an object is dropped from the aircraft. What is

posledela

Explanation:

use the diagram^

horizontal velocity of the object nvr changes, it'll stay 50m/s till it hits the ground

initial vertical velocity = 0

to find the final vertical velocity, use the formula:

$v {}^{2} = u {}^{2} + 2as$

where u = 0m/s, a = 9.8m/s², and s = 2000m

solve for v.

for time taken to reach ground, use the formula:

$v = u + at$

where v = the value for v u calculated above, u = 0 m/s, a = 9.8m/s² and solve for t.

7 0

3 years ago

More than 5 points and will pick brainliest answer plus I will give thanks❗️❕❗️(Don’t answer if you don’t know the answer)

Setler79 [48]

The answer is 2 and 3 because light bulbs dont produce uv rays and uv rays are very dangerous to people.

5 0

4 years ago

4.A balloon containing an ideal gas sits in a constant pressure chamber. The temperature of the chamber is raised from 275 K to

Natalka [10]

To solve this problem, we must take into account the equation:

$PV=nRT$

So, for an isobaric transformation (constant pressure), an increase in volume generates an increase in temperature. This can be written as:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$ $\frac{0.03}{275}=\frac{V_2}{375}\Rightarrow V_2=\frac{0.03*375}{275}=0.04091m^3$

Thus, the new volume will be 0.04091m^3

8 0

1 year ago