Answer:
If the force applied is larger than 185.2 N, yes.
Explanation:
In order to move the table, the pushing force must be larger than the frictional force. The frictional force is given by:
where
is the coefficient of static friction
is the mass of the table
is the gravitational acceleration
Substituting,
So, we are able to move the table if we push with a force larger than 185.2 N.
A manufacturer supplies plastic cups that are placed under the legs of the chair. The manufacturer's claim is that there will be no marks of chair due to pushing or pulling on the floor.
<h3>Who is a manufacturer?</h3>
A manufacturer is a person who converts his idea of developing and constructing any item using some set of machines.
A manufacturer supplies plastic cups that are placed under the legs of the chair.
When there is a relative motion between two objects, friction force causes the opposition to this motion. Between chair and floor, there will be high magnitude of force. But if we insert plastic cups in the legs of chair, then the friction between cup and floor will be less. This keeps the floor protected form any scratches.
Learn more about manufacturer.
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Answer:
(a) Time t = 16.46 sec
(b) Time t =13.466 sec
(c) Deceleration = 
Explanation:
(a) As the train starts from rest its initial velocity u = 0 m/sec
Acceleration 
Final speed v = 80 km/hr
From first equation of motion v =u+at
So 
(b) Now initial speed u = 22.22 m/sec
As finally train comes to rest so final speed v=0 m/sec
Deceleration 
So 
(c) We have given that initial velocity = 80 km/hr = 22.22 m/sec
Final velocity v = 0 m/sec
Time t = 8.30 sec
So acceleration is given by
As acceleration is negative so it is a deceleration
Complete question:
A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?
Answer:
the weight of the air is 76.44 lbs
Explanation:
Given;
dimension of the dormitory, = 14 ft by 13 ft by 6 ft
density of the air, = 0.07 lbs/ft³
The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft
= 1092 ft³
The weight of the air = density x volume
= 0.07 lbs/ft³ x 1092 ft³
= 76.44 lbs
Therefore, the weight of the air is 76.44 lbs
Answer:
4.91 x 10⁻⁷ m
Explanation:
the applicable formula is
v = fλ
where
v = velocity (i.e speed) = given as 3.0 x 10⁸ m/s
f = frequency = given asw 6.11 x 10¹⁴
λ = wavelength
if we rearrange the equation and substitute the values given above,
v = fλ
λ = v/f
= 3.0 x 10⁸ / 6.11 x 10¹⁴
= 4.91 x 10⁻⁷ m
