Answer:
μ = 0.6
Explanation:
F = μN
N = mg
F = μmg
3 N = μ*0.5 kg * 9.8 m/s²
μ = 3/(0.5*9.8) = 0.6
Answer: 552.5 watts
Explanation:
Given that,
Weight of man = 956 N
Height (h) = 2.41 m
Time taken (t) = 4.17 seconds
Power = ?
Recall that power is the rate of work done per unit time i.e Power = work/time
Thus, power = (mgh) / t
(Since weight = mass x acceleration due to gravity)
Power = (weight x height)/ time
Power = (956N x 2.41 m) / 4.17s
Power = 2303.96/4.17
Power = 552.5 watts
Thus, 552.5 watts of power is generated.
Answer:
Let's say the pitcher is angry or just has a really heavy hand while throwing this ball, and now you have to catch it, otherwise it's going to ram into your face. When you put your hands up just in time to catch this ball, this is called impulse, or commonly expressed as a reflex. Depending on what kind of ball is being thrown, such as a golf ball, baseball, basketball, beach-ball, rubber-ball, baseball, etc. ... the weight of the ball itself is going to impact how much it i going to hurt when you catch it without any hand protection. However, if you're catching, let's say a baseball, with a padded glove, it is not going to hurt as bad as catching the baseball bare handed, because the padded glove has enough padding in it to create a barrier between the skin of your hand and the palm of the glove.
well the formula to find speed is distance divided by time and you already have the time so all you need to know to find the average speed of the car is the total distance around the racetrake.
Answer:
1.84 m
Explanation:
For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.
So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,
a = g + rω²
= 9.8 m/s² + 1.10 m × (18.85 rad/s)²
= 9.8 m/s² + 390.85 m/s²
= 400.65 m/s²
Now, using v² = u² + 2a(h₂ - h₁) where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.
v² = u² + 2a(h₂ - h₁)
So, v² - u² = 2a(h₂ - h₁)
h₂ - h₁ = (v² - u²)/2a
h₂ = h₁ + (v² - u²)/2a
substituting the values of the variables into the equation, we have
h₂ = 1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)
h₂ = 1.3 m + [-430.15 (m/s)²]/-801.3 m/s²
h₂ = 1.3 m + 0.54 m
h₂ = 1.84 m
