WILL GIVE BRAINLIST: Does an objects weight affect how fast it falls?

A 850-kg sports car accelerates from rest to 95 km/h in 6.8 s .

swat32

From the calculations, the power expended is 43650 W.

<h3>What is the power expended?</h3>

Now we can find the acceleration from;

v = u + at

u = 0 m/s

v = 95 km/h or 26.4 m/s

t = 6.8 s

a = ?

Now

v = at

a = v/t

a = 26.4 m/s/ 6.8 s

a = 3.88 m/s^2

Force = ma = 850-kg * 3.88 m/s^2 = 3298 N

The distance covered is obtained from;

v^2 = u^2 + 2as

v^2 = 2as

s = v^2/2a

s = (26.4)^2/2 * 3.88

s = 696.96/7.76

s = 90 m

Now;

Work = Fs

Work = 3298 N * 90 m = 296820 J

Power = 296820 J/ 6.8 s

= 43650 W

Learn more about power expended:brainly.com/question/11579192

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5 0

2 years ago

If a 0.08 kg cell phone falls off a table at 15 m/s, then what is its kinetic energy right before it hits the ground?

Mariana [72]

The kinetic energy of the phone right before it hits the ground is 9J.

<h3>Kinetic energy of the phone</h3>

The kinetic energy of the phone right before it hits the ground is calculated as follows;

K.E = ¹/₂mv²

where;

m is mass of the phone
v is velocity of the phone

K.E = ¹/₂(0.08)(15)²

K.E = 9 J

Thus, the kinetic energy of the phone right before it hits the ground is 9J.

Learn more about kinetic energy here: brainly.com/question/25959744

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7 0

1 year ago

To increase the current in a circuit, which can be increased?

wariber [46]

Answer:

A. Voltage

Explanation:

I got a 100% on the quiz on edg so trust me

4 0

3 years ago

A biologist looking through a microscope sees a bacterium at r⃗ 1=2.2i^+3.7j^−1.2k^μm(1μm=10−6m). After 6.2 s , it's at r⃗ 2=4.6

stiv31 [10]

The Average velocity for the bacterium is 0.75 unit/sec.

<u>Explanation:</u>

The given values are in the vector form

Where,

dS = distance covered

dT = time interval

Now, to calculate distance covered, we have

$|d S|=\sqrt{d S^{2}}$

&

$d S=r_{2}-r_{1}$

d S=(4.6 i+1.9 k)-(2.2 i+3.7 j - 1.2 k)

d S=(4.6-2.2) i+(0-3.7) j+(1.9+1.2) k

d S=2.4 i-3.7 j+3.1 k

Now, putting these values in the standard formula to evaluate the average velocity, we get;

$v_{a v g}=\frac{|\mathrm{d} S|}{d T}$

$v(a v g)=\frac{|\sqrt{\left\{\left(2.4^{2}\right)+\left(3.7^{2}\right)+\left(3.1^{2}\right)\right\}}|}{7.2}$

As dT=7.2 sec

Now,

Solving the equation, we get;

$v(a v g)=\frac{5.390732789}{7.2}$

$\begin{aligned}&v(a v g)=\frac{5.39}{7.2}\\&v(a v g)=0.748611111\\&v(a v g)=0.75 \text { units / sec }\end{aligned}$

Hence, the average velocity for the bacterium is 0.75 unit/sec.

3 0

3 years ago

You are on a new planet, and are trying to use a simple pendulum to find the gravity experienced on this new planet. You find th

levacccp [35]

To solve this problem it is necessary to apply the concepts related to the Period based on the length of its rope and gravity, mathematically it can be expressed as

$T= 2\pi \sqrt{\frac{L}{g}}$

g = Gravity

L = Length

T = Period

Re-arrange to find the gravity we have

$g = \frac{4\pi^2 L}{T^2}$

Our values are given as

$L = 0.35m\\T = 2s\\$

Replacing we have

$g = \frac{4\pi^2 L}{T^2}$

$g = \frac{4\pi^2 0.35}{2^2}$

$g = 3.45 m/s^2$

Therefore the correct answer is C.

4 0

3 years ago