Answer:
The horizontal component of her velocity is approximately 1.389 m/s
The vertical component of her velocity is approximately 7.878 m/s
Explanation:
The given question parameters are;
The initial velocity with which Margaret leaps, v = 8.0 m/s
The angle to the horizontal with which she jumps, θ = 80° to the horizontal
The horizontal component of her velocity, vₓ = v × cos(θ)
∴ vₓ = 8.0 × cos(80°) ≈ 1.389
The horizontal component of her velocity, vₓ ≈ 1.389 m/s
The vertical component of her velocity, 
= v × sin(θ)
∴ = 8.0 × sin(80°) ≈ 7.878
The vertical component of her velocity, ≈ 7.878 m/s.
For a 50 kg person receives an absorbed dose of gamma radiation of 20 millirads, the total energy absorbed is mathematically given as
E=0.1457J
<h3>What is the total energy absorbed?</h3>
Generally, the equation for the total energy absorbed is mathematically given as
E=mass*gamma radiation
Therefore
E=50*20*19^{-3}
E=0.1457J
In conclusion, the total energy absorbed
E=0.1457J
Read more about Energy
brainly.com/question/13439286
Answer:
The distance is
=
7
m
Explanation:
Apply the equation of motion
s
(
t
)
=
u
t
+
1
2
a
t
2
The initial velocity is
u
=
0
m
s
−
1
The acceleration is
a
=
2
m
s
−
2
Therefore, when
t
=
3
s
, we get
s
(
3
)
=
0
+
1
2
⋅
2
⋅
3
2
=
9
m
and when
t
=
4
s
s
(
4
)
=
0
+
1
2
⋅
2
⋅
4
2
=
16
m
Therefore,
The distance travelled in the fourth second is
d
=
s
(
4
)
−
s
(
3
)
=
16
−
9
=
7
m
Answer:
Explanation:
Given that,
The instantaneous current in the circuit is giveen by :
We need to find the rms value of the current.
The general equation of current is given by :
It means, 
We know that,
So, the rms value of current is 2.12 A.
Answer:
Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.
Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.
Explanation:
Part a
When Road is Level
The stopping sight distance is given as
Here
- SSD is the stopping sight distance which is to be calculated.
- u is the speed which is given as 60 mi/hr
- t is the perception-reaction time given as 2.5 sec.
- a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
- G is the grade of the road, which is this case is 0 as the road is level
Substituting values
So the minimum stopping sight distance is 563.36 ft.
Part b
When Road has a maximum grade of 4%
The stopping sight distance is given as
Here
- SSD is the stopping sight distance which is to be calculated.
- u is the speed which is given as 60 mi/hr
- t is the perception-reaction time given as 2.5 sec.
- a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
- G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade
For upgrade of 4%, Substituting values
<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>
For downgrade of 4%, Substituting values
<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>
As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.
