The efforts applied are 320Nm.
What is work or effort?
Work is when we apply force on body and displacement occur are called work .
W= Fs
where F is force and s is displacement .
displacement is 2-1.6
. = 0.4m
By applying the value in formula
We have given F =800N
W= 800× 0.4
W= 320Nm.
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The peak radiation will occur at 97 nm.
Answer:
3.53 second
Explanation:
The formula for the height is
When it hits the ground, the height is zero.
So, put h = 0 in the above equation
Take positive sign
t = 3.53 second.
Thus, the time taken to hit the ground is 3.53 second.
Answer:
Answer is 20
Explanation:
As we know
\Large\boxed{ \tt{}F =ma}
F=ma
Here Given
★ F = 100
★ m = 25 kg
Putting in the equation
\leadsto\tt{100=25\times\:a}⇝100=25×a
On solving
\sf{\dashrightarrow\:a = 4 \:ms^{-2}}⇢a=4ms
−2
Now the body initially is stationary therefore initial velocity (u) must be zero
u = 0
Applying Newton 1st Law of Motion :
\sf\Large\boxed{ \tt{}v =u + at}
v=u+at
Substituting the values to find final velocity at t= 5 sec
\leadsto \tt{v = 0 + 4 \times \:5}⇝v=0+4×5
\tt{ \pink{\dashrightarrow\:v = 20 \:m/s}}⇢v=20m/s
☞ Hence, the final velocity is 20 m/s
Walking at a speed of 2.1 m/s, in the first 2 s John would have walked
(2.1 m/s) (2 s) = 4.2 m
Take this point in time to be the starting point. Then John's distance from the starting line at time <em>t</em> after the first 2 s is
<em>J(t)</em> = 4.2 m + (2.1 m/s) <em>t</em>
while Ryan's position is
<em>R(t)</em> = 100 m - (1.8 m/s) <em>t</em>
where Ryan's velocity is negative because he is moving in the opposite direction.
(b) Solve for the time when they meet. This happens when <em>J(t)</em> = <em>R(t)</em> :
4.2 m + (2.1 m/s) <em>t</em> = 100 m - (1.8 m/s) <em>t</em>
(2.1 m/s) <em>t</em> + (1.8 m/s) <em>t</em> = 100 m - 4.2 m
(3.9 m/s) <em>t</em> = 95.8 m
<em>t</em> = (95.8 m) / (3.9 m/s) ≈ 24.6 s
(a) Evaluate either <em>J(t)</em> or <em>R(t)</em> at the time from part (b).
<em>J</em> (24.6 s) = 4.2 m + (2.1 m/s) (24.6 s) ≈ 55.8 m