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

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

We can solve the problem by using Ohm's law, which states that an Ohmic conductor the following relationship holds:

where

is the potential difference applied to the resistor
I is the current flowing through it
R is the resistance
In our problem, I=4.00 A and

, so the potential difference is
The frequency of oscillation is 2.153 Hz
What is the frequency of spring?
Spring Frequency is the natural frequency of spring with a weight at the lower end. Spring is fixed from the upper end and the lower end is free.
For the mass-spring system in this problem,
The Frequency of spring is calculated with the equation:

Where,
f = frequency of spring
k = spring constant = 64 N/m
m = mass attached to spring = 350g = 0.350 kg
a = maximum acceleration = 5.3 m/s^2
Substituting the values in the equation,



Hence,
The frequency of oscillation is 2.153 Hz
Learn more about frequency here:
<u>brainly.com/question/13978015</u>
#SPJ4
Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

W = 3.266 N
The mass of the meters stick is :

So, the mass of the meter stick is 0.333 kg.
I'm going to say false hope that helped