Answer:
Formation of new elements
Explanation:
Answer: 2, the nuclear strong force drops to practically nothing at large distances.
Explanation: The protons and neutrons in the nucleus share subatomic particles called pions. This exchange is what keeps the protons and neutrons stuck together in the nucleus. Despite the strong force being the strongest force, it has a very small range. This is because pions have very short lifespans. So, the strong force would have literally no effect at large distances.
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<h2>When two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of Q. The mass of P is half that of Q. The ratio of the specific heat capacity of P to Q</h2>
Explanation:
Specific heat capacity
It is defined as amount of heat required to raise the temperature of a substance by one degree celsius .
It is given as :
Heat absorbed = mass of substance x specific heat capacity x rise in temperature
or ,
Q= m x c x t
In above question , it is given :
For Q
mass of Q = m
Temperature changed =T₂/2
Heat supplied = x
Q= mc t
or
X=m x C₁ X T₁
or, X =m x C₁ x T₂/2
or, C₁=X x 2 /m x T₂ (equation 1 )
For another quantity : P
mass of P =m/2
Temperature= T₂
Heat supplied is same that is : X
so, X= m/2 x C₂ x T₂
or, C₂=2X/m. T₂ (equation 2 )
Now taking ratio of C₂ to c₁, We have
C₂/C₁= 2X /m.T₂ /2X /m.T₂
so, C₂/C₁= 1/1
so, the ratio is 1: 1
Answer:
There are three ways an object can accelerate: a change in velocity, a change in direction, or a change in both velocity and direction.
Explanation:
Answer:
solution:
to find the speed of a jogger use the following relation:
V
=
d
x
/d
t
=
7.5
×m
i
/
h
r
...........................(
1
)
in Above equation in x and t. Separating the variables and integrating,
∫
d
x
/7.5
×=
∫
d
t
+
C
or
−
4.7619
=
t
+
C
Here C =constant of integration.
x
=
0 at t
=
0
, we get: C
=
−
4.7619
now we have the relation to find the position and time for the jogger as:
−
4.7619 =
t
−
4.7619
.
.
.
.
.
.
.
.
.
(
2
)
Here
x is measured in miles and t in hours.
(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),
to get:
= −
4.7619
=
1
−
4.7619
= −
3.7619
or x
=
7.15
m
i
l
e
s
(b) To find the jogger's acceleration in m
i
l
/
differentiate
equation (1) with respect to time.
we have to eliminate x from the equation (1) using equation (2).
Eliminating x we get:
v
=
7.5×
Now differentiating above equation w.r.t time we get:
a
=
d
v/
d
t
=
−
0.675
/
At
t
=
0
the joggers acceleration is :
a
=
−
0.675
m
i
l
/
=
−
4.34
×
f
t
/
(c) required time for the jogger to run 6 miles is obtained by setting
x
=
6 in equation (2). We get:
−
4.7619
(
1
−
(
0.04
×
6 )
)^
7
/
10=
t
−
4.7619
or
t
=
0.832
h
r
s
