Answer:
E = 120.77 J
Explanation:
Given that,
Mass of a baseball, m = 0.143 kg
Velocity of a baseball, v = 41.1 m/s
We need to find the kinetic energy of the baseball. We know that the kinetic energy of an object is associated with its motion. It can be given by the formula as follow :
So, the kinetic energy of the baseball is 120.77 J.
The apparent velocity is B) 48 m/s north
Explanation:
Here we have a problem of relativity of velocities.
In fact, the train is travelling north at a speed of
where this velocity is measured with respect to the ground.
At the same time, a passenger on the train is walking towards the rear (so, south) at a velocity of
where this velocity is measured with respect to the train, which is in motion in the opposite direction.
Therefore, the apparent velocity of the passenger with respect to an observer standing on the ground is:
And the direction is north, since this number is positive.
Learn more about velocity:
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Answer:
3.75 × 10⁻⁸ N
Explanation:
Given:
Intensity of the electromagnetic wave, I = 150 W/m²
Sides of the board = 25 cm (= 0.25 m) and 30 cm (= 0.30 m)
therefore,
the area of the rectangular box, A = 0.25 × 0.30 = 0.075 m²
Now,
force exerted on the card by the radiation, F =
here,
C is the speed of the light = 3 × 10⁸ m/s
on substituting the respective values, we get
F =
or
F = 3.75 × 10⁻⁸ N
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45 )
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = - 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct