Answer:
The speed of the flow of water in the river is 4 m/s.
Explanation:
Given that,
Speed of man = 5 m/s
If this man swims crosses a specific river his speed is 3 m/s.
If he takes the minimum time to cross the river
Let the speed of flow of water be
We need to calculate the speed of the flow of water in the river
Using formula for velocity
Where, = velocity of swimmer
= relative velocity
= velocity of river
Put the value into the formula
Hence, The speed of the flow of water in the river is 4 m/s.
Answer:
The box will be moving at 0.45m/s. The solution to this problem requires the knowledge and application of newtons second law of motion and the knowledge of linear motion. The vertical component of the force Fp acts vertically upwards against the directio of motion. This causes a constant upward force of 23sin45° to act on the box. Fhe frictional force of 13N also acts vertically upwards and so two forces act upwards against rhe force of gravity resulting un a net force of 0.7N acting kn the box. This corresponds to an acceleration of 0.225m/s². So in w.0s after i start to push v = 0.45m/s.
Explanation:
Answer : Resistance is 16 ohms.
Explanation :
It is given that,
Power of toaster,
Voltage,
According to ohm's law, V = I R
Mathematically power is defined as :
or
So, the resistance of the toaster is 16 ohms.
Hence, this is the required solution.
Answer:
= 3.87 m/s
Explanation:
Since momentum is conserved, momentum right before and after the boy jumps onto the skateboard should be the same.
Initial momentum = momentum of boy = 43.2 x 4.10 = 177 kg·m/s
Final momentum = momentum of boy and skateboard = (43.2 + 2.50) x v = 45.7v
177 = 45.7v
v = 177 / 45.7 = 3.87 m/s
(Hope this helps can I pls have brainlist (crown)☺️)
Answer:
Explanation:
Length = 1.00 m
If the length is 1.0, the vertical distance pivot to bob is cos 35 = 0.819
At the lowest point, vertical distance is 1.0, so the change is the difference, 0.181 meter
The potential energy of that height is converted to kinetic energy of motion, which determines the speed.
PE = KE
mgh = ½mV²
V = √(2gh) = 1.88 m/s