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2 years ago

B) If the actual counterweight fitted to this boom was

Physics

1 answer:

Lostsunrise [7]2 years ago

6 0

Answer:

50 N

4.2 N

Explanation:

i) The force needed to balance the boom is 2400 N. If the weight of the counterbalance is 2350 N, then the downward force the park attendant must apply is 50 N.

ii) When the boom is resting on the end support, the normal force is:

∑τ = Iα

-W (0.50) + F (3.0) − N (6.0) = 0

-0.50 W + 3.0 F = 6.0 N

N = (-0.50 W + 3.0 F) / 6.0

N = (-0.50 × 2350 + 3.0 × 400) / 6.0

N ≈ 4.2

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A 2.3 kg block of copper is heated at atmospheric pressure such that its temperature increases from 6 oC to 90 oC. How much heat

Shtirlitz [24]

Answer:

the heat absorbed by the block of copper is 74368.476J

Explanation:

Hello!

To solve this problem use the first law of thermodynamics that states that the heat applied to a system is the difference between the initial and final energy considering that the mass and the specific heat do not change so we can infer the following equation

Q=mCp(T2-T1)

Where

Q=heat

m=mass=2.3kg

Cp=0.092 kcal/(kg C)=384.93J/kgK

T2=Final temperatura= 90C

T1= initial temperature=6 C

solving

$Q=(2.3kg)(384.93\frac{J}{kgC} )(90C-6C)=74368.476J$

the heat absorbed by the block of copper is 74368.476J

7 0

2 years ago

the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force

-Dominant- [34]

Answer:

PART A)

External force will be 75 N

PART B)

distance moved will be 1.125 m

Explanation:

PART A)

Given that net force on the mower is

$F_{net} = 51 N$

now we also know that friction force due to ground is given as

$F_f = 24 N$

now we have

$F_{net} = F_{ext} - F_f$

$51 = F_{ext} - 24$

$F_{ext} = 75 N$

so external force will be 75 N

PART B)

deceleration due to friction when external force is removed from it

$a = \frac{F_f}{m}$

$a = \frac{24}{24} = 1 m/s^2$

now we can find the distance by kinematics

$v_f^2 - v_i^2 = 2 a d$

$0 - 1.5^2 = 2(-1)d$

$d = 1.125 m$

so the distance moved will be 1.125 m

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2 years ago

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. how much time does it ta

Yuri [45]

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. The time taken by the car will be 4 seconds. the correct answer is option(c).

When acceleration is constant, the rate of change in velocity is also constant. In the absence of any acceleration, velocity remains constant. When acceleration is positive, velocity becomes more significant.

Let a denote acceleration, u denote initial velocity, v denote final velocity, and t denote time.

The equation of motion is stated as,

v = u + at

v² = u² + 2as

A car travels across a distance of 60.0 m while accelerating constantly from 12.0 m/s to 18.0 m/s.

Then the time taken by the car will be,

u = 12 m/s

v = 18 m/s

s = 60 m

Put these in the equation v² = u² + 2as.

18² = 12² + 2 x a x 60

a = 1.5

Then the time will be

18 = 12 + 1.5t

1.5t = 6

t = 4 seconds

Hence, the time taken is 4 s.

The complete question is:

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. How much time does it take?

1.00 s

2.50 s

4.00 s

4.50 s

To know more about acceleration refer to: brainly.com/question/12550364

#SPJ4

8 0

1 year ago

Your roommate leaves a 120W fan running in your apartment.Over the course of an hour,how much thermal energy does the fan add to

zubka84 [21]

Answer:

$4.32\cdot 10^5 J$

Explanation:

Power is related to energy by the following relationship:

$P=\frac{E}{t}$

where

P is the power used

E is the energy used

t is the time elapsed

In this problem, we know that

- the power of the fan is P = 120 W

- the fan has been running for one hour, which corresponds to a time of

$t = 1 h \cdot (60 min/h)(60 s/min)=3600 s$

So we can re-arrange the previous equation to find E, the energy (in the form of thermal energy) released by the fan:

$E=Pt=(120 W)(3600 s)=4.32\cdot 10^5 J$

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75 percent (calculated percentage %) of what number equals 27? Answer: 36.

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