Hey There,
Question: "<span>A student gives a brief push to a block of dry ice. A moment later, the block moves across a very smooth surface at a constant speed. When drawing the free body diagram for the block of dry ice moving at a constant speed, the forces that should be included are: (select all that apply)"
Answer: C. Force Of Friction
B. Force
If This Helps May I Have Brainliest?</span>
Answer:
Explanation:
The electric field outside the sphere is given as,
E = k Q /r²
here Q = n x 1.6 x 10⁻¹⁹ C
where n is the number of electons
if the dimeter of sphere d= 25 cm= 0.25 m
then the radius r = 0.125 m
we get
n= E r²/ k x 1.6 x 10⁻¹⁹ C
n = 1350N/C x (0.125m)² / (8.99 x 10⁹ N m²/C² x 1.6 x 10⁻¹⁹ C)
n = 14664731646
Answer:
a = 0.7267
, acceleration is positive therefore the speed is increasing
Explanation:
The definition of acceleration is
a = dv / dt
they give us the function of speed
v = - (t-1) sin (t² / 2)
a = - sin (t²/2) - (t-1) cos (t²/2) 2t / 2
a = - sin (t²/2) - t (t-1) cos (t²/2)
the acceleration for t = 4 s
a = - sin (4²/2) - 4 (4-1) cos (4²/2)
a = -sin 8 - 12 cos 8
remember that the angles are in radians
a = 0.7267
the problem does not indicate the units, but to be correct they must be m/s²
We see that the acceleration is positive therefore the speed is increasing
Answer:
4.535 N.m
Explanation:
To solve this question, we're going to use the formula for moment of inertia
I = mL²/12
Where
I = moment of inertia
m = mass of the ladder, 7.98 kg
L = length of the ladder, 4.15 m
On solving we have
I = 7.98 * (4.15)² / 12
I = (7.98 * 17.2225) / 12
I = 137.44 / 12
I = 11.45 kg·m²
That is the moment of inertia about the center.
Using this moment of inertia, we multiply it by the angular acceleration to get the needed torque. So that
τ = 11.453 kg·m² * 0.395 rad/s²
τ = 4.535 N·m
Answer:
the final velocity of the car is 59.33 m/s [N]
Explanation:
Given;
acceleration of the car, a = 13 m/s²
initial velocity of the car, u = 120 km/h = 33.33 m/s
duration of the car motion, t = 2 s
The final velocity of the car in the same direction is calculated as follows;
v = u + at
where;
v is the final velocity of the car
v = 33.33 + (13 x 2)
v = 59.33 m/s [N]
Therefore, the final velocity of the car is 59.33 m/s [N]
