3 years ago

Help please due by 3:00

Physics

1 answer:

FrozenT [24]3 years ago

8 0

Answer:

zero

Explanation:

Acceleration is a measure of the rate of change of velocity. If the velocity is unchanging, its rate of change is zero.

The acceleration is zero.

You might be interested in

How much power is needed to lift a 20-kg object to a height of 4.0 m in 2.5 seconds?

4vir4ik [10]

Answer:

313.92w

Explanation:

Formula for power:

P=W/∆t = Fv

Givens:

m=20kg

∆y=4.0m

∆t=2.5s

a=9.81m/s²

In order to find power, we first need to solve for work.

W=Fd (force*displacement), f=mg

W=mg∆y

W=(20kg)(9.81m/s²)(4.0m)

W=784.8J

P=W/∆t

P=784.8J/2.5s

P=313.92 watts

5 0

2 years ago

Question 3 of 10

AlexFokin [52]

The answer is Jupiter

6 0

2 years ago

Read 2 more answers

develop a plan to test the hardness of a sample of feldspar using the following items glass plate copper penny and streak plate

Artemon [7]

Stroke then 3883 you choke <3333333333

6 0

2 years ago

Stick a fork into each end of the root vegetable as shown. The forks should be on the same side of the vegetable, and

zhuklara [117]

Answer:

Yes

Explanation:

There is a position that works better than this and that is switching the sides of the forks.

7 0

2 years ago

Daring Darless wishes to cross the Grand Canyon of the Snake River by being shot from a cannon. She wishes to be launched at 65

guajiro [1.7K]

Answer:

She must be launched with minimum speed of <u>57.67 m/s</u> to clear the 520 m gap.

Step-by-step explanation:

Given:

The angle of projection of the projectile is, $\theta =65$ °

Range of the projectile is, $R=520$ m.

Acceleration due to gravity, $g=9.8\ m/s^2$

The minimum speed to cross the gap is the initial speed of the projectile and can be determined using the formula for range of projectile.

The range of projectile is given as:

$R=\frac{v_{0}^2\sin2\theta}{g}$

Plug in all the given values and solve for minimum speed, $v_0$ .

$520=\frac{v_{0}^2\sin(2(65))}{9.8}\\520\times 9.8=v_{0}^2\sin(130)\\5096=1.532v_{0}^2\\v_0^2=\frac{5096}{1.532}\\v_0^2=3326.371\\v_0=\sqrt{3326.371}=57.67\textrm{ m/s}$

Therefore, she must be launched with minimum speed of 57.67 m/s to clear the 520 m gap.

3 0

3 years ago