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Sunny_sXe [5.5K]
3 years ago
7

True or False: It would be easier to run on a planet with high gravity than one with less gravity.

Physics
1 answer:
LUCKY_DIMON [66]3 years ago
3 0
True! It would be easier
You might be interested in
A 235 kg crate is pulled across a horizontal surface with a force of 760 N applied at an
AnnyKZ [126]

Answer:

658.16N

Explanation:

Step one:

given data

mass m= 235kg

Force F= 760N

angle= 30 degrees

Required

The horizontal component of the force

Step two:

The horizontal component of the force

Fh= 760cos∅

Fh=760cos30

Fh=760*0.8660

Fh=658.16N

3 0
3 years ago
The third one pls help
Nadya [2.5K]

Answer:

20 ms¯¹

Explanation:

3. Determination of the final velocity

From the question given above, the following data were obtained:

Time (t) = 4 s

Acceleration (a) = 5 ms¯²

Initial velocity (u) = 0 ms¯¹

Final velocity (v) =?

Acceleration is simply defined as the change in velocity per unit time.

Mathematically, it can be expressed as:

Acceleration (a) = final velocity – Initial velocity / time

a = v – u / t

With the above formula, we can obtain the final velocity of the car as follow:

Time (t) = 4 s

Acceleration (a) = 5 ms¯²

Initial velocity (u) = 0 ms¯¹

Final velocity (v) =?

a = v – u / t

5 = v – 0 / 4

5 = v / 4

Cross multiply

v = 5 × 4

v = 20 ms¯¹

Thus, the final velocity of the car is 20 ms¯¹

7 0
3 years ago
An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e
VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

r = \frac{8.26}{2} = 4.13m

And the distance 'd' is

d = lsin\theta

d = 1.14 sin 16.2\°

d = 0.318m

Through the free-body diagram the tension components are given by

Tcos\theta = mg

Tsin\theta = \frac{mv^2}{R}

Here we can watch that,

R = r+d

Dividing both expression we have that,

tan\theta = \frac{v^2}{Rg}

Replacing the values,

tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}

v = 4.83371m/s

PART B) Using the vertical component we can find the tension,

Tcos\theta = mg

T = \frac{mg}{cos\theta}

T = \frac{(10+26.2)(9.8)}{cos(16.2)}

T = 369.42N

6 0
3 years ago
Help!!! If anyone could do one of these, i'm confused on how I should write the equation down.
balu736 [363]

F=ma=m(change in velocity/change in time)

Number 1

F=ma

F=55kg(1.1ms^-1/1.6s)=37.8N

Number 2

F=ma

F=0.440kg(10ms^-1/0.02s)=220N

Number 3

F=ma

F=1400kg(15ms^-1/0.73s)=2.88*10^3N or 28,767N

Any questions please feel free to ask.

4 0
3 years ago
Caffeine concentration is 1.99 mg/oz how many cans would be leathal if 10g was leathal and there where 12oz in a can
katovenus [111]

The number of cans that would be considered lethal if 10g was lethal and there where 12oz in a can is 419 cans.

<h3>How to convert mass?</h3>

According to this question, caffeine concentration is 1.99 mg/oz.

1.99 milligrams can be converted to grams as follows:

1.99milligrams ÷ 1000 = 0.00199grams

This means that 0.00199grams per oz is the caffeine concentration.

If there were 12 oz in a can, then, 0.00199grams × 12 = 0.02388 grams in 1 can.

This means that if 10grams is considered lethal, 10grams ÷ 0.02388 grams = 419 cans would be lethal for consumption.

Therefore, the number of cans that would be considered lethal if 10g was lethal and there where 12oz in a can is 419 cans.

Learn more about conversion factor at: brainly.com/question/14479308

#SPJ1

5 0
1 year ago
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