On Earth, the gravitational field strength is 10 N/kg. Calculate Ep for a 4 kg bowling ball have that is being held 2 m above th
1 answer:
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The coefficient of friction is 0.051
Explanation:
The motion of the skater is a uniformly accelerated motion, therefore we can use the following suvat equation:
where:
v = 0 is the final velocity of the skater (he comes to a stop)
u = 10.0 m/s is his initial velocity
a is the acceleration
is the distance he travels before stopping
Solving for a, we find the acceleration of the skater:
We also know that the net force acting on the skater is the force of friction, therefore we can write (Newton's second law of motion):
where
is the force of friction
m is the mass of the skater
is the coefficient of friction
is the acceleration
is the acceleration of gravity
Solving for , we find the coefficient of friction:
Learn more about friction:
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Answer:
Explanation:
First of all we shall calculate electric field near charge 2Q .
electric field due to charge Q = K x Q / (5 x 10⁻² )²
E₁ = KQ / 25 x 10⁻⁴ = KQ x 10⁴ / 25 . It is acting along positive x axis
E₁ = KQ x 10⁴ i / 25
Similarly electric field due to charge 3Q near 2Q
= 3KQ x 10⁴ i / 25 . It is acting along y-axis
E₂ = 3KQ x 10⁴ j / 25
Similarly electric field due to charge 4Q near 2Q
= 4KQ x 10⁴ j / (25 x 2 )
= 2 KQ x 10⁴ / 25 . It is acting acting along north east direction
unit vector in north east direction = ( i + j )/ √2
So E₃ can be represented by
E₃ = 2 KQ x 10⁴ ( i + j ) / 25 x √2
Total field = KQ x 10⁴ i / 25 + 3KQ x 10⁴ j / 25 + 2 KQ x 10⁴ ( i + j ) / ( 25 x √2 )
= KQ x 10⁴ [ i + 3 j + √2 i + √2 j ) / 25
= 400 KQ ( 2.414 i + 4.414 j ) N / C
Force on 2Q = Field x charge = 400 KQ ( 2.414 i + 4.414 j ) x 2Q N
= 800 KQ² ( 2.414 i + 4.414 j ) N
= 800 x 9 x 10⁹ x ( 2.5 x 10⁻⁶ )² x 2.414 x ( i + 2 j ) N
= 108.63 ( i + 2 j ) N .
Magnitude of this force
= 108.63 x √5
= 243 N approx .
Answer:
1/18
Explanation:
The number of ways in which two die can be thrown is the sample space of the experiment
(1,1), (1,2) (1,3), (1,4) (1,5) (1,6)
(2,1), (2,2) (2,3), (2,4) (2,5) (2,6)
(3,1), (3,2) (3,3), (3,4) (3,5) (3,6)
(4,1), (4,2) (4,3), (4,4) (4,5) (4,6)
(5,1), (5,2) (5,3), (5,4) (5,5) (5,6)
(6,1), (6,2) (6,3), (6,4) (6,5) (6,6)
From here it can be seen that there are only two cases where the sum on the two die is 11 i.e., (6,5) and (5,6)
∴ Probability of getting 11 is<u> 1/18</u>