Answer:
a. 
b.
must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping
c.
is the time taken to stop after braking
Explanation:
Given:
- speed of leading car,

- speed of lagging car,

- distance between the cars,

- deceleration of the leading car after braking,

a.
Time taken by the car to stop:

where:
, final velocity after braking
time taken


b.
using the eq. of motion for the given condition:

where:
final velocity of the chasing car after braking = 0
acceleration of the chasing car after braking

must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping
c.
time taken by the chasing car to stop:


is the time taken to stop after braking
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
Answer:
W ≅ 292.97 J
Explanation:
1)What is the work done by tension before the block goes up the incline? (On the horizontal surface.)
Workdone by the tension before the block goes up the incline on the horizontal surface can be calculated using the expression;
W = (Fcosθ)d
Given that:
Tension of the force = 62 N
angle of incline θ = 34°
distance d =5.7 m.
Then;
W = 62 × cos(34) × 5.7
W = 353.4 cos(34)
W = 353.4 × 0.8290
W = 292.9686 J
W ≅ 292.97 J
Hence, the work done by tension before the block goes up the incline = 292.97 J
Answer:
Mass of the steel cube = 7800 kg
Volume of the steel = 1.025 cubic centimetre
Explanation:
Given:
The density of the steel = 7.8
Side of the cube = 12 cm
<u>(1)The mass of steel cube :</u>
We know that,

We are given with density and sides of the cube
then volume of the cube
=
= 
= 1000 cubic centimetre
Now


mass = 7800 kg
<u>(2)volume of steel:</u>
Given the mass = 8 kg

Substituting the values


volume = 1.025 cubic centimetre