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2 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]2 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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2. If a rock fell down a cliff and hit the bottom of the ravine at 4 seconds, how fast was the rock

Iteru [2.4K]

Answer: -39.2 m/s or 39.2 m/s directed downwards

Explanation:

This situation is a good example of Free Fall, where the main condition is that the initial velocity must be zero $V_{o}=0$ , and the acceleration is constant (acceleration due gravity).

So, in order to calculate the final velocity $V$ of the rock just at the moment it hitsthe bottom of the cliff, we will use the following equation:

$V={V_{o}}^{2}+gt$

Where:

$g=-9.8 m/s^{2}$ is the acceleration due gravity (directed downwards)

$t=4 s$ is the time it takes to the rock to fall down the cliff

$V=(-9.8 m/s^{2})(4 s)$

$V=-39.2 m/s$ This is the rock's final velocity and its negative sign indicates it is directed downwards

6 0

3 years ago

Read 2 more answers

In a face, you run 3000 meters east in 21 minutes. What is your velocity in km/min?

Anni [7]

$3 \times \ 21$

7 0

2 years ago

I’m not sure how to do 30, could someone pls help?

choli [55]

I think its C or D.

Einstein's Nobel winning photoelectric equation.

The higher the work function, the more energy needed to photoeject an electron from the surfface

Try D

8 0

2 years ago

An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounce

Ierofanga [76]

Answer:

2.67 m

Explanation:

k = Spring constant = 1.5 N/m

g = Acceleration due to gravity = 9.81 m/s²

l = Unstretched length

Frequency of SHM motion is given by

$f_s=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

Frequency of pendulum is given by

$f_p=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$

Given in the question

$f_p=\dfrac{1}{2}f_s$

$\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow \sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{g}{l}=\dfrac{1}{4}\dfrac{k}{m}\\\Rightarrow l=\dfrac{4gm}{k}\\\Rightarrow l=\dfrac{4\times 9.81\times \dfrac{1}{9.81}}{1.5}\\\Rightarrow l=2.67\ m$

The unstretched length of the spring is 2.67 m

6 0

2 years ago

A particle with a mass of 6.64 × 10–27 kg and a charge of +3.20 × 10–19 C is accelerated from rest through a potential differenc

blondinia [14]

Answer:

Explanation:

Given that,

Mass m = 6.64×10^-27kg

Charge q = 3.2×10^-19C

Potential difference V =2.45×10^6V

Magnetic field B =1.6T

The force in a magnetic field is given as Force = q•(V×B)

Since V and B are perpendicular i.e 90°

Force =q•V•BSin90

F=q•V•B

So we need to find the velocity

Then, K•E is equal to work done by charge I.e K•E=U

K•E =½mV²

K•E =½ ×6.64×10^-27 V²

K•E = 3.32×10^-27 V²

U = q•V

U = 3.2×10^-19 × 2.45×10^6

U =7.84×10^-13

Then, K•E = U

3.32×10^-27V² = 7.84×10^-13

V² = 7.84×10^-13 / 3.32×10^-27

V² = 2.36×10^14

V=√2.36×10^14

V = 1.537×10^7 m/s

So, applying this to force in magnetic field

F=q•V•B

F= 3.2×10^-19 × 1.537×10^7 ×1.6

F = 7.87×10^-12 N

6 0

2 years ago

Read 2 more answers