Answer:
an ice cream machine produced 44 ice cream per minute.
Now, after reconditioning it speeds up to 55 ice cream per minute.
=> 55 - 44 = 11
=> 11 / 44 = 0.25
=> 0.25 * 100% = 25%.
thus it increased it production to 25%.
Explanation:
Answer:
d = 2,042 10-3 m
Explanation:
The laser diffracts in the circular slit, so the process equation is
d sin θ= m λ
The first diffraction minimum occurs for m = 1
We can use trigonometry in the mirror
tan θ = Y / L
Where L is the distance from the Moon to Earth
Since the angle is extremely small
tan θ = sin θ / cos θ
Cos θ = 1
tant θ = sin θ = y / L
We replace
d y / L = λ
d = λ L / y
Let's calculate
d = 532 10⁻⁹ 3.84 10⁶/1 10³
d = 2,042 10-3 m
The kinematic equations of motion that apply here are<span>y(t)=votsin(θ)−12gt2</span>and<span>x(t)=votcos(θ)</span>Setting y(t)=0 yields <span>0=votsin(θ)−12gt2</span>. If we solve for t, we obtain, by factoring,<span>t=<span>2vsin(θ)g</span></span>Substitute this into our equation for x(t). This yields<span>x(t)=<span><span>2v2cos(θ)sin(θ)</span>g</span></span><span>This is equal to x=<span><span>v^2sin(2θ)</span>g</span></span>Hence the angles that have identical projectiles are have the same range via substitution in the last equation is C. <span> 60.23°, 29.77° </span>
Answer:
F n = 0.2 N
Explanation:
given,
you are exerting force of 10 N on the ball.
mass of the ball = 1 kg
acceleration due to gravity = 9.8 m/s²
normal force on the ball = ?
normal force is force exerted by the object to counteract the force from other object.
normal force acting on the ball will be
F n = F - mg
F n = 10 - 1 × 9.8
F n = 10 -9.8
F n = 0.2 N
Hence, normal force acting on the ball is equal to 0.2 N
Answer:
The speed of space station floor is 49.49 m/s.
Explanation:
Given that,
Mass of astronaut = 56 kg
Radius = 250 m
We need to calculate the speed of space station floor
Using centripetal force and newton's second law




Where, v = speed of space station floor
r = radius
g = acceleration due to gravity
Put the value into the formula


Hence, The speed of space station floor is 49.49 m/s.