Answer:
The slower runner is 1.71 km from the finish line when the fastest runner finishes the race.
Explanation:
Given;
the speed of the slower runner, u₁ = 11.8 km/hr
the speed of the fastest runner, u₂ = 15 km/hr
distance, d = 8 km
The time when the fastest runner finishes the race is given by;
The distance covered by the slower runner at this time is given by;
d₁ = u₁ x 0.533 hr
d₁ = 11.8 km/hr x 0.533 hr
d₁ = 6.29 km
Additional distance (x) the slower runner need to finish is given by;
6.29 km + x = 8km
x = 8 k m - 6.29 km
x = 1.71 km
Therefore, the slower runner is 1.71 km from the finish line when the fastest runner finishes the race.
Answer:
θ=π/2
Explanation:
The definition of work is W = → F ⋅ → d = q E c o s θ d W=F→⋅d→=qEcosθd. So if no work is done, the displacement must be in the direction perpendicular to the force ie c o s θ = 0 → θ = π / 2 cosθ=0→θ=π/2
A charged particle can be displaced without any external work done on it in a uniform electric field when its movement is perpendicular to the direction of the electric field.
In a uniform electric field, the electric force is the same in every direction. Therefore, if a charge were to be displaced perpendicular to the original direction of the electric field (i.e., in the y or z direction), it would not encounter any extra electric forces. This means there would be no external work being done on the charge. When a charge is moved perpendicular to an electric field, the field does not affect it, and hence, no work is done by the field.
In other words, a charge can be displaced in this field without any external work being done on it when it is moved in a direction perpendicular to the uniform electric field, either in y-axis or z-axis, assuming the electric field is constant in the x-axis direction.
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0.67 m/s2
0.075 m/s2
54 m/s2
height at which the ball was hit is 3.0 m tall, how far did the ball go horizontally
before it hit the ground?
5.5 m
3.6 m
O 4.3 m
4.2 m
Answer:
5.5 is the correct answer
please keeps as Brainly list
Answer:
The total percent cold work done is 36.46%
Explanation:
Let initial metal thickness = T
Final metal thickness = t
The percent cold work done = WC
Then
%Wc = (T - t)/T × 100
% Wc = ( 0.096 - 0.061 )/0.096 ×100
Total %WC = 36.46%
Answer:
The total percent of cold work is 57.34%
Explanation:
Let x the initial thickness of the sheet. After 33% of cold working, the thickness is 0.096 in. Then:
x - 0.33x = 0.096
x = 0.143 in
the final thickness is equal to 0.061 in. The percent of cold work done is:
%
The angular speed should be 17.18 rad / s
Since
moment of inertia of the disc I = 1/2 m r²
= .5 x 8 x .25²
= .25 kg m²
Now the work done by force should be converted into the rotational kinetic energy
F x d = 1/2 I ω²
here,
F is the force applied,
d is displacement,
I is moment of inertia of disc
and ω is angular velocity of disc
So,
41 x .9 = 1/2 x .25 ω²
ω² = .25
ω = 17.18 rad / s
Learn more about speed here: brainly.com/question/18742396
Answer:
Explanation:
moment of inertia of the disc I = 1/2 m r²
= .5 x 8 x .25²
= .25 kg m²
The wok done by force will be converted into rotational kinetic energy
F x d = 1/2 I ω²
F is force applied , d is displacement , I is moment of inertia of disc and ω
is angular velocity of disc
41 x .9 = 1/2 x .25 ω²
ω² = .25
ω = 17.18 rad / s
B. Sprouted bean seeds increase.
C. Sprouted bean seeds remain constant.
D. None of the above
As we go up the y-axis, the number of sproutedbean seeds increase (option B).
Graph is a data chart intended to illustrate the relationship between a set (or sets) of numbers (quantities, measurements or indicative numbers) and a reference set.
In a graph, there are two axes as follows;
According to this question, a graph of number of sprouted bean seeds on the y-axis is plotted against temperature on the x-axis.
We can observe that as we go up the y-axis, the number of sprouted bean seeds increase.
Learn more about graphs at: brainly.com/question/2938738
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