true or false when taking a handoff you need to make sure you you reach out with your hands and grab the ball

The boom of a fire truck raises a fireman (and his equipment – total weight 280 lbf) 60 ft into the air to fight a building fire

Katen [24]

Answer:

a)Work done by fireman= 2.15 Btu

b) Time t= 0.86 sec

Explanation:

Given that

Weight = 280 lbf

We know that 1 lbf = 4.44 N

so 280 lbf = 1245.5 N

Weight =1245.5 N

Height h = 60 ft

We know that

1 ft = 0.3048 m

So 60 ft = 18.28 m

h =18.28 m

Power = 3.5 hp

We know that

1 hp =0.74 KW

So 3.5 hp = 2.61 KW

Power = 2.61 KJ/s

So the work done by fireman = Weight x h

Now by putting the values

Work done by fireman= 1245.5 x 18.28 J

Work done by fireman= 2267.74 J

Work done by fireman= 2.26774 KJ

We know that 1 Btu= 1.05 KJ

So 2.266 KJ = 2.15 Btu

Work done by fireman= 2.15 Btu

We know that ,rate of work is called power.

Power x time = work

2.61 x t = 2.26

So t= 0.86 sec

6 0

3 years ago

How many main retrofit strategies help solve the ecological problems with suburbs?

iragen [17]

ANSWER:

5

Explanation:

Because they are elven in numbers

3 0

3 years ago

In an experiment, a disk is set into motion such that it rotates with a constant angular speed. As the disk spins, a small spher

boyakko [2]

Answer:

L₀ = L_f , K_f < K₀

Explanation:

For this exercise we start as the angular momentum, with the friction force they are negligible and if we define the system as formed by the disk and the clay sphere, the forces during the collision are internal and therefore the angular momentum is conserved.

This means that the angular momentum before and after the collision changes.

Initial instant. Before the crash

L₀ = I₀ w₀

Final moment. Right after the crash

L_f = (I₀ + mr²) w

we treat the clay sphere as a point particle

how the angular momentum is conserved

L₀ = L_f

I₀ w₀ = (I₀ + mr²) w

w = $\frac{I_o}{I_o + m r^2}$ w₀

having the angular velocities we can calculate the kinetic energy

starting point. Before the crash

K₀ = ½ I₀ w₀²

final point. After the crash

K_f = ½ (I₀ + mr²) w²

sustitute

K_f = ½ (I₀ + mr²) ( $\frac{I_o}{I_o + m r^2}$ w₀)²

Kf = ½ $\frac{I_o^2}{ I_o + m r^2}$ w₀²

we look for the relationship between the kinetic energy

$\frac{K_f}{K_o}$ = $\frac{I_o}{I_o + m r^2}$

$\frac{K_f}{K_o } < 1$

K_f < K₀

we see that the kinetic energy is not constant in the process, this implies that part of the energy is transformed into potential energy during the collision

6 0

3 years ago

1. A negatively charged rod is moved near the top of a positively charged electroscope. What

Xelga [282]

Answer:

A) Moves closer together

7 0

3 years ago

Two blocks of masses 3 kg and 5 kg approach each other with initial velocities 4 m/s and -6 m/s respectively. The two blocks col

Kipish [7]

Answer:

Explanation:

After the collision velocity of the particle is (4î - 3ĵ)m/s . ... A particle of mass 1 kg moving with a velocity of (4i^−3j^)m/s collides with a fixed surface. ... Perfectly inelastic. D ... The common velocity of the blocks after collision is: ... A ball falls from a height of 5 m and strikes the roof of a lift. ... Stay upto date with our Newsletter! i know this is not right but just here for points see ya loser

6 0

3 years ago

true or false when taking a handoff you need to make sure you you reach out with your hands and grab the ball ​

true or false when taking a handoff you need to make sure you you reach out with your hands and grab the ball