Answer:
t = 0.2845Nm (rounded to 4 decimal places)
Explanation:
The disk rotates at a distance of an arc length of 28cm
Arc length = radius × central angle × π/180
28cm = 10cm × central angle × π/180
Central angle = 
× 180/π ≈ 160.4°
Torque (t) = rFsin(central angle) , where F is the applied force
Radius in meters = 10/100 = 0.1m
t = 0.1m × 16N × sin160.4°
t = 0.2845Nm (rounded to 4 decimal places)
Answer:
Explanation:
frequency of whistle = 1.85 x 10² = 185 Hz
frequency of beat heard = 8 beat /s . No of beat produced is equal to difference of frequencies of two sound source . Here difference is created due to Doppler effect . One of the train is moving so it will have apparent frequency which is different one from its original frequency .
When the moving train is approaching the observer , its frequency will be higher . As beat is heard at the rate of 8 beats / s , apparent frequency of approaching train will be 185 + 8 = 193 Hz .
Applying Doppler's formula of apparent frequency ,
193 = 185 x V / ( V - v ) , where V is velocity of sound and v is velocity of train .
193 V - 193 v = 185 V
193 v = 8 V
v = 8 x V / 193
= 8 x 343 / 193
= 14.21 m /s
Second possibility is that apparent velocity is less ie 185 - 8 = 177 Hz
In that case moving train will be moving away from observer . If its velocity be v
177 = 185 x V / ( V + v )
177 V + 177 v = 185 V
v = 8 x 343 / 177
= 15.50 m /s .
Answer:
The coefficient of static friction between your partner and the floor is 0.55
Explanation:
Given:
Mass 
Kg
Frictional force 
N
From the formula of frictional force,
Where 
coefficient of static friction, 
Put the above values and find the coefficient of static friction.
Therefore, the coefficient of static friction between your partner and the floor is 0.55
Answer:
The capacitor having less distance of separation has a stronger electric field.
Explanation:
The capacitors are identical and only difference between them is that one has twice the plate separation of the other. Therefore, capacitance of the given capacitors C1 and C2 is,
C1= Aε/d and C2=Aε/2d
The charges Q1 and Q2 on the capacitors of capacitance C1 and C2 respectively, is then given by the equation,
Q1=VC1
Q1=VAε/d
Q2=VC2
Q2=VAε/2d
Therefore, the surface charge density σ1 and σ2 for the capacitors is,
σ1=Q1/A
σ1=VAε/(d*A)
σ1=Vε/d
Similarly,
σ2=Q2/A
σ2=Vε/2d
The electric field between the plates is directly proportional to the surface charge density. And so electric field is inversely proportional to the distance of separation. Therefore the capacitor whose distance of separation is less has a stronger electric field.
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