as we know that
acceleration = 0.50 m/s^2
distance traveled = 125 m
time taken = 20 s
now we will use the equation of kinematics
![d = v_i*t + \frac{1}{2}at^2](https://tex.z-dn.net/?f=d%20%3D%20v_i%2At%20%2B%20%5Cfrac%7B1%7D%7B2%7Dat%5E2)
![125 = v_i* 20 + \frac{1}{2}*0.50*20^2](https://tex.z-dn.net/?f=125%20%3D%20v_i%2A%2020%20%2B%20%5Cfrac%7B1%7D%7B2%7D%2A0.50%2A20%5E2)
![125 = 20*v_i + 100](https://tex.z-dn.net/?f=125%20%3D%2020%2Av_i%20%2B%20100)
![125 - 100 = 20*v_i](https://tex.z-dn.net/?f=125%20-%20100%20%3D%2020%2Av_i)
![v_i = 1.25 m/s](https://tex.z-dn.net/?f=v_i%20%3D%201.25%20m%2Fs)
so its initial speed must be 1.25 m/s
Answer:
Kinetic energy increases because iron shavings move in the direction of magnetic force.
Potential energy decreases, and kinetic and thermal energy increase.
Potential energy increases because movement is in the opposite direction of the magnetic field.
They seek a rail line not affected by friction.
Answer:
High specific heat -> takes more energy to raise/lower object's temperature
Low specific heat -> takes less energy to raise/lower object's temperature
Explanation:
The specific heat capacity is the amount of heat required to raise the temperature of something per unit of mass.
A high specific heat value for an object means it takes more energy to raise or lower that object's temperature. A low specific heat value for an object means it does not take very much energy to heat or cool that object.
Answer:
the mass of the cart is 150 kg
Explanation:
given,
mass of boy(m) = 50 kg
speed of boy (v)= 10 m/s
initial velocity of cart (u) = 0
final velocity of cart(V) = 2.5 m/s
mass of the cart(M) = ?
m v + M u = (m + M ) V......................(1)
50× 10 + 0 = (50 + M ) 2.5
M =![\dfrac{500}{2.5} - 50](https://tex.z-dn.net/?f=%5Cdfrac%7B500%7D%7B2.5%7D%20-%2050)
M = 150 Kg
hence, the mass of the cart is 150 kg
Complete question:
while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8.32 meters per second take the speed of sound as 340 meters per second. calculate the frequency reflected off the wall to the bat?
Answer:
The frequency reflected by the stationary wall to the bat is 41 kHz
Explanation:
Given;
frequency emitted by the bat, = 39 kHz
velocity of the bat,
= 8.32 m/s
speed of sound in air, v = 340 m/s
The apparent frequency of sound striking the wall is calculated as;
![f' = f(\frac{v}{v- v_b} )\\\\f' = 39,000(\frac{340}{340 -8.32} )\\\\f' = 39978.29 \ Hz](https://tex.z-dn.net/?f=f%27%20%3D%20f%28%5Cfrac%7Bv%7D%7Bv-%20v_b%7D%20%29%5C%5C%5C%5Cf%27%20%3D%2039%2C000%28%5Cfrac%7B340%7D%7B340%20-8.32%7D%20%29%5C%5C%5C%5Cf%27%20%3D%2039978.29%20%5C%20Hz)
The frequency reflected by the stationary wall to the bat is calculated as;
![f_s = f'(\frac{v + v_b}{v} )\\\\f_s = 39978.29(\frac{340 + 8.32}{340} )\\\\f_s = 40,956.56 \ Hz](https://tex.z-dn.net/?f=f_s%20%3D%20f%27%28%5Cfrac%7Bv%20%2B%20v_b%7D%7Bv%7D%20%29%5C%5C%5C%5Cf_s%20%3D%2039978.29%28%5Cfrac%7B340%20%2B%208.32%7D%7B340%7D%20%29%5C%5C%5C%5Cf_s%20%3D%2040%2C956.56%20%5C%20Hz)
![f_s\approx 41 \ kHz](https://tex.z-dn.net/?f=f_s%5Capprox%2041%20%5C%20kHz)