Answer:
Force exerted, F = 1.5 N
Explanation:
It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.
i.e. u = 0
v = 30 m/s
Time taken, t = 0.06 s
Mass of the paper, m = 0.003 kg
We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :



F = 1.5 N
So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.
Answer:
c=0.14J/gC
Explanation:
A.
2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.
B.
We can use the expression for heat transmission

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

for water we have to
c = 4.18J / g ° C
replacing we have

I hope this is useful for you
A.
2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.
B.
Podemos usar la expresión para la transmisión de calor

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

para el agua tenemos que
c=4.18J/g°C
reemplazando tenemos

Answer:
c
Explanation:
cuz its informing the length of 5 and weight on 20N
Answer:
²₁H + ³₂He —> ⁴₂He + ¹₁H
Explanation:
From the question given above,
²₁H + ³₂He —> __ + ¹₁H
Let ⁿₐX be the unknown.
Thus the equation becomes:
²₁H + ³₂He —> ⁿₐX + ¹₁H
We shall determine, n, a and X. This can be obtained as follow:
For n:
2 + 3 = n + 1
5 = n + 1
Collect like terms
n = 5 – 1
n = 4
For a:
1 + 2 = a + 1
3 = a + 1
Collect like terms
a = 3 – 1
a = 2
For X:
n = 4
a = 2
X =?
ⁿₐX => ⁴₂X => ⁴₂He
Thus, the balanced equation is
²₁H + ³₂He —> ⁴₂He + ¹₁H