Answer:
No, it is not necessary for them to have same mass.
Explanation:
Let both bodies have a density d1 and d2 respectively.
Since their volumes are equal V1 = V2
we know that, https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7Bvolume%7D
Hence, d1 = and d2 =
Taking the ratio of densities,we get
This implies that unless the bodies have same densities, the mass of the two bodies will not be same.
F=9/5C+32 is the formula for Celsius to Fahrenheit.
Answer:
9) This is a case of deceleration
10)-0.8 ms-2
b) acceleration is the change in velocity with time
11)
a) 100 ms-1
b) 100 seconds
12) 10ms-1
13) more information is needed to answer the question
14) - 0.4 ms^-2
15) 0.8 ms^-2
Explanation:
The deceleration is;
v-u/t
v= final velocity
u= initial velocity
t= time taken
20-60/50 =- 40/50= -0.8 ms-2
11)
Since it starts from rest, u=0 hence
v= u + at
v= 10 ×10
v= 100 ms-1
b)
v= u + at but u=0
1000 = 10 t
t= 1000/10
t= 100 seconds
12) since the sprinter must have started from rest, u= 0
v= u + at
v= 5 × 2
v= 10ms-1
14)
v- u/t
10 - 20/ 25
10/25
=- 0.4 ms^-2
15)
a=v-u/t
From rest, u=0
8 - 0/10
a= 8/10
a= 0.8 ms^-2
All the radiation from stars is the result of nuclear fusion in their cores.
Answer:
lowest frequency = 535.93 Hz
distance between adjacent anti nodes is 4.25 cm
Explanation:
given data
length L = 32 cm = 0.32 m
to find out
frequency and distance between adjacent anti nodes
solution
we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s
so
lowest frequency will be = 
..............1
put here value in equation 1
lowest frequency will be = 
lowest frequency = 535.93 Hz
and
we have given highest frequency f = 4000Hz
so
wavelength = 
..............2
put here value
wavelength = 
wavelength = 0.08575 m
so distance = 
..............3
distance = 
distance = 0.0425 m
so distance between adjacent anti nodes is 4.25 cm
