Answer:
<h3>B. 19miles</h3>
Explanation:
If Freddy drives 4 miles east to his friend's house. He then travels 9 more miles east to the supermarket. Finally on his way back home he out of gas 6 miles after leaving the supermarket, the distance travel by fred will be the sup of all the distances he covered throughout the journey.
Distance covered by fred = 4miles + 9miles + 6miles
Distance covered by fred = 13miles + 6miles
Distance covered by fred = 19miles
Answer:
5.0 m/s
Explanation:
The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

where u is the initial speed and
. The horizontal distance travelled by the salmon is

where d = 1.95 m and t is the time needed to reach the final point.
Re-arranging for t,
(1)
Along the vertical direction, the equation of motion is

where:
y = 0.311 m is the final height reached by the salmon
h = 0 is the initial height
is the vertical component of the initial velocity of the salmon
is the acceleration of gravity
t is the time
Substituting t as found in eq.(1), we get the equation

and we can solve this formula for u, the initial speed of the salmon:

Answer:

Explanation:
Given that,
The distance between two spheres, r = 25 cm = 0.25 m
The capacitance, C = 26 pF = 26×10⁻¹² F
Charge, Q = 12 nC = 12 × 10⁻⁹ C
We need to find the work done in moving the charge. We know that, work done is given by :

Put all the values,

So, the work done is
.
Answer:
Explanation:
We shall apply conservation of momentum law in vector form to solve the problem .
Initial momentum = 0
momentum of 12 g piece
= .012 x 37 i since it moves along x axis .
= .444 i
momentum of 22 g
= .022 x 34 j
= .748 j
Let momentum of third piece = p
total momentum
= p + .444 i + .748 j
so
applying conservation law of momentum
p + .444 i + .748 j = 0
p = - .444 i - .748 j
magnitude of p
= √ ( .444² + .748² )
= .87 kg m /s
mass of third piece = 58 - ( 12 + 22 )
= 24 g = .024 kg
if v be its velocity
.024 v = .87
v = 36.25 m / s .
Answer:
4452.5 J.
Explanation:
The diver have both kinetic and potential energy.
Ek = 1/2mv² ................. Equation 1
Where Ek = Kinetic Energy of the diver, m = mass of the diver, v = velocity of the diver.
Given: m = 65 kg, v = 6.4 m/s.
Substitute into equation 1
Ek = 1/2(65)(6.4²)
Ek = 1331.2 J.
Also,
Ep = mgh ............................ Equation 2
Where Ep = Potential energy of the diver when its above the water, h = height of the diver above the water, g = acceleration due to gravity.
Given: m = 65 kg, h = 4.9 m, g = 9.8 m/s²
Substitute into equation 2.
Ep = 65(4.9)(9.8)
Ep = 3121.3 J.
Note: When she hits the water, the potential energy is converted to kinetic energy.
E = Ek+Ep
Where E = Kinetic energy of the diver when she hits the water.
E = 1331.2+3121.3
E = 4452.5 J.