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2 years ago

A caterpillar crawls 20 feet in 5 minutes. What is the caterpillars crawling speed?

Physics

1 answer:

Pavel [41]2 years ago

6 0

The equation is
s= d/t

In this case you would have to write it out as:
s= 20/5
Speed = 4

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A uniform string of length 10.0 m and weight 0.32 N is attached to the ceiling. A weight of 1.00 kN hangs from its lower end. Th

Juliette [100K]

Answer: 0.0180701 s

Explanation:

Given the following :

Length of string (L) = 10 m

Weight of string (W) = 0.32 N

Weight attached to lower end = 1kN = 1×10^3

Using the relation:

Time (t) = √ (weight of string * Length) / weight attached to lower end * acceleration due to gravity

g = acceleration due to gravity = 9.8m/s^2

Weight of string = 0.32N

Time(t) = √ (0.32 * 10) / [(1*10^3) * (9.8)]

Time = √3.2 / 9800

= √0.0003265

= 0.0180701s

5 0

2 years ago

There are Z protons in the nucleus of an atom, where Z is the atomic number of the element. An α particle carries a charge of +2

qaws [65]

Answer:

$r = 2.84 \times 10^{-14} m$

Explanation:

As per energy conservation we know that the electrostatic potential energy of the charge system is equal to the initial kinetic energy of the alpha particle

So here we can write it as

$\frac{1}{2}mv^2 = \frac{k(2e)(ze)}{r}$

now we know that

$m = 1.67 \times 10^{-27} kg$

$e = 1.6 \times 10^{-19} C$

z = 79

here kinetic energy of the incident alpha particle is given as

$KE = 6.4 \times 10^{-13} J$

now we have

$6.4 \times 10^{-13} = \frac{(9\times 10^9)(1.6 \times 10^{-19})^2(79)}{r}$

now we have

$r = 2.84 \times 10^{-14} m$

7 0

3 years ago

A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3° angle. What i

Oksi-84 [34.3K]

Answer:

v_{1fy} = - 0.4549 m / s

Explanation:

6 0

2 years ago

3. If I run 150m in 30 seconds, what speed will I have been running at?

Radda [10]

Answer:

speed = distance/time

Explanation:

speed = 150/30

speed =5m/s

you were running fast .....5m/s is a good speed

7 0

2 years ago

A- 1000 m/s2<br> Xi-0m<br> Xf-0.75m<br> Vf-?

sleet_krkn [62]

Answer:

The final velocity of the object is, $v_{f}$ = 27 m/s

Explanation:

Given,

The acceleration of the object, a = 1000 m/s²

The initial displacement of the object, $x_{i}$ = 0 m

The final displacement of the object, $x_{f}$ = 0.75 m

The initial velocity of the object will be, $v_{i}$ = o m/s

The final velocity of the object, $v_{f}$ = ?

The average velocity of the object,

v = ( $x_{f}$ - $x_{i}$ )/ t

= 0.75 / t

The acceleration is given by the relation

a = v / t

1000 m/s² = 0.75 / t²

t² = 7.5 x 10⁻⁴

t = 0.027 s

Using the I equation of motion,

$v_{f}$ = u + at

Substituting the values

$v_{f}$ = 0 + 1000 x 0.027

= 27 m/s

Hence, the final velocity of the object is, $v_{f}$ = 27 m/s

8 0

3 years ago