Answer:
Explanation:
Given
Wheels are rotating with constant angular velocity let say 
Presence of constant angular velocity show that there is no angular acceleration thus there is no tangential acceleration.
But any particle on the rim will experience a constant acceleration towards center called centripetal acceleration.
(a) yes, there will be tangential velocity which is given by
where r=radial distance from center
(b)tangential acceleration
there would be no tangential acceleration as velocity is constant
(c)centripetal acceleration
Yes, there will be centripetal acceleration given by
Answer:
3.5m/s^2
Explanation:
From Newton's second Law of Motion
F = ma
Where F is the applied force, m is the mass of the object and a is the acceleration.
F = 350 N
Mass = 100kg
350N = 100×a
a = 350/100
a = 3.5m/s^2
The acceleration of the object will be 3.5m/s^2
Answer:
41.74 m/s
Explanation:
The energy used to draw the bowstring = the kinetic energy of the arrow.
Fd = 1/2mv²................................ Equation 1
Where F = force, d = distance move string, m = mass of the arrow, v = speed of the arrow.
make v the subject of the equation
v = √(2Fd/m)...................... Equation 2
Given: F = 201 N, m = 0.3 kg, d = 1.3 m.
Substitute into equation 2
v = √(2×201×1.3/0.3)
v = √(1742)
v = 41.74 m/s.
Hence the arrow leave the bow with a speed of 41.74 m/s
Answer:
Option A
Explanation:
The statement makes sense since it's already explained that the galaxy is moving away from us and unlike option C which depicts that the galaxy is moving to us.
This statement makes sense. The redshift means that we see the galaxy moving away from us, so observers in that galaxy must also see us moving away from them—which means they see us redshifted as well
The electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.
To determine σ:
σ = Q/A
Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:
σ = Q/d²
Make this substitution in the equation for E:
E = Q/(2ε₀d²)
We see that E is inversely proportional to the square of d:
E ∝ 1/d²
The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude:
