A girl exerts a horizontal force of 109 N on a crate with a mass of 31.2 kg. HINT (a) If the crate doesn't move, what's the magn
1 answer:
Answer:
(a) Magnitude of static friction force is 109 N
(b) Minimum possible value of static friction is 0.356
Solution:
As per the question;
Horizontal force exerted by the girl, F = 109 N
Mass of the crate, m = 31.2 kg
Now,
(a) To calculate the magnitude of static friction force:
Since, the crate is at rest, the forces on the crate are balanced and thus the horizontal force is equal to the frictional force, f:
F = f = 109 N
(b) The maximum possible force of friction between the floor and the crate is given by:
where
N = Normal reaction = mg
Thus
For the crate to remain at rest, The force exerted on the crate must be less than or equal to the maximum force of friction.
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Answer:
Explanation:
Given:
Capacitance, C = 85 pF = 85 × 10⁻¹² F
Resistance, R = 75 MΩ = 75×10⁶Ω
Charge in capacitor at any time 't' is given as:
where,
Q₀ = Maximum charge = CE
E = Initial voltage
t = time
also, Q = CV
V= Final voltage = 90% of E = 0.9E
thus, we have
or
or
taking log both sides, we get
or
or
or
Answer:
The magnitude of the force F is given by
F = ( +
+
) *(
*g/(
))
Explanation:
Given there are three blocks of masses ,
and
(ref image in attachment)
When all three masses move together at an acceleration a, the force F is given by
F = ( +
+
) *a ................(equation 1)
Also it is given that does not move with respect to
, which gives tension T is exerted on pulley by
only, Hence tension T is
T = *a ..........(equation 2)
There is also also tension exerted by . There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by
T =
................(equation 3)
From equation 2 and 3, we get
*a =
Squaring both sides we get
*
=
* (
+
)
*
= (
*
)+ (
*
)
( -
) *
=
*
=
*
/(
-
)
Taking square root on both sides, we get acceleration a
a = *g/(
)
Hence substituting the value of a in equation 1, we get
F = ( +
+
) *(
*g/(
))
Answer:
E = 8.5 * 10^6 V/m
Explanation:
In general we have the following relation between the Electric Field and the Elecric Potential:
Due to the vector nature of the electric filed, we can only know the mean Electric field E across the membrane, and take it out from the integral, that is:
E = (ΔV)/L
Where L is the thickness of the membrane and ΔV is the potential difference.
Therefore:
E = 8.53933*10^6 V/m
rounding to the first tenth:
E = 8.5 * 10^6 V/m