Answer:
to jamal then luis then to eva then back to owen
A bodybuilder deadlifts 215 kg to a height of 0.90 m. If he deadlifts this weight 10 times in 45 s, the power exerted is 421 W (b.)
<h3>What is power?</h3>
In physics, power (P) is the work (W) done over a period of time.
- Step 1. Calculate the work done by the bodybuilder each time.
The bodybuilder lifts a 215 kg (m) weight to a height of 0.90 m (h). Being the gravity (g) of 9.81 m/s², we can calculate the work done in each lift using the following expression.
W = m × g × h = 215 kg × 9.81 m/s² × 0.90 m = 1.9 × 10³ N
- Step 2. Calculate the work done by the bodybuilder over 10 times.
W = 10 × 1.9 × 10³ N = 1.9 × 10⁴ N
- Step 3. Calculate the power exerted by the bodybuilder.
The bodybuilder does a work of 1.9 × 10⁴ N in a 45-s span.
P = 1.9 × 10⁴ N/45 s = 421 W
A bodybuilder deadlifts 215 kg to a height of 0.90 m. If he deadlifts this weight 10 times in 45 s, the power exerted is 421 W (b.)
Learn more about power here: brainly.com/question/911620
#SPJ1
Your answer is B
Plz give brainliest and I will answer all your future questions
Answer:
v = 21 m / s
Explanation:
We can solve this exercise with the kinematics equations, let's start by finding the acceleration of the train with the initial data
v = v₀ + a t
the initial speed is the speed within the city 6 m / s, the final speed is v = 11 m / s and the time is t = 8 s
a = (v-v₀) / t
a = (11 - 6) / 8
a = 0.625 m / s²
when it leaves the city with speed vo = 11 m / s it accelerates for t = 16 s
v = v₀ + a t
v = 11 + 0.625 16
v = 21 m / s
The speed of the second mass after it has moved ℎ=2.47 meters will be 1.09 m/s approximately
<h3>
What are we to consider in equilibrium ?</h3>
Whenever the friction in the pulley is negligible, the two blocks will accelerate at the same magnitude. Also, the tension at both sides will be the same.
Given that a large mass m1=5.75 kg and is attached to a smaller mass m2=3.53 kg by a string and the mass of the pulley and string are negligible compared to the other two masses. Mass 1 is started with an initial downward speed of 2.13 m/s.
The acceleration at which they will both move will be;
a = (
- 
) / ( + )
a = (5.75 - 3.53) / (5.75 + 3.53)
a = 2.22 / 9.28
a = 0.24 m/s²
Let us assume that the second mass starts from rest, and the distance covered is the h = 2.47 m
We can use third equation of motion to calculate the speed of mass 2 after it has moved ℎ=2.47 meters.
v² = u² + 2as
since u =0
v² = 2 × 0.24 × 2.47
v² = 1.1856
v = √1.19
v = 1.0888 m/s
Therefore, the speed of mass 2 after it has moved ℎ=2.47 meters will be 1.09 m/s approximately
Learn more about Equilibrium here: brainly.com/question/517289
#SPJ1
