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3 years ago

How far did a frog jump if he travels at a rate of 2.1 m/s for 10 seconds?

Physics

1 answer:

Anestetic [448]3 years ago

3 0

Answer:

21 m

Explanation:

The motion of the frog is a uniform motion (constant speed), therefore we can find the distance travelled by using

$d=vt$

where

d is the distance covered

v is the speed

t is the time

The frog in this problem has a speed of

v = 2.1 m/s

and therefore, after t = 10 s, the distance it covered is

$d=(2.1)(10)=21 m$

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Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

4 0

2 years ago

A space station in the shape of a 100 m-diameter (50m radius) wheel is spinning so as to impart a linear tangential speed of 22.

zalisa [80]

Answer:

correct option is b. 31.3 m/s

Explanation:

given data

artificial gravity a1 = 1 g

artificial gravity a2 = 2 g

diameter = 100 m

radius r= 50 m

speed v1 = 22.1 m/s

solution

As acceleration is ∝ v²

so we can say

$\frac{a2}{a1} = \frac{v2}{v1}$ .....................1

put here value

$\frac{2}{1} = \frac{v2}{22.1}$

solve it

v2 = $\sqrt{2 }$ × 22.1

v2 = 31.25 m/s

so correct option is b. 31.3 m/s

4 0

2 years ago

How would I find the average distance for this?

Maslowich

Answer:

Explanation:add them then divide them by 100 I think

3 0

2 years ago

Read 2 more answers

a hippopotamus produces a pressure of 250000 pa when it is standing on all four feet if the weight of the hippo is 40000 N what

mamaluj [8]

0.04m²

Explanation:

Given parameters:

Pressure = 250000Pa

Weight = 40000N

Unknown:

Area of each foot = ?

Solution:

Pressure is the force exerted per unit area of a body

Pressure = $\frac{force}{area}$

To find the area;

Area = $\frac{force }{pressure}$

Area = $\frac{40000}{250000}$ = 0.16m²

The force exerted by all the four feet is 0.16m²

the area of each feet = $\frac{0.16}{4}$ = 0.04m²

Learn more:

Pressure brainly.com/question/7139767

#learnwithBrainly

8 0

2 years ago

Michael is doing an experiment. He wants to drop a bowling ball and a stuffed bear out the window of a tall building. Under what

Anon25 [30]

Answer:

If there was no air resistance

Explanation:

We know that free fall is a unique motion in which gravity only works on one object. Objects that are said to be free-falling do not experience a significant force of air resistance; They come under the sole effect of gravity. Under such conditions, all objects fall under the same acceleration, regardless of their mass.

6 0

2 years ago