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3 years ago

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you are riding an elevator upwards at 2.20m/s. you have a balance accurately calibrated in newtons. when you hang a mass of 15.0

the balance, the reading of the balance is

1 answer:

Nady [450]3 years ago

5 0

Answer:

<h2> 33N</h2>

Explanation:

Note: the unit for acceleration in m/s^2

Also, the unit of the mass in the problem is assumed to be in kg

We are going to use newtons second law of motion which expresses the weight in an elevator as

F=ma

Given data

Acceleration of the elevator a= 2.2m/s^2

mass m = 15kg

substituting the data in the expression for force is

F=2.2*15

F= 33N

The reading of the balance is 33N

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Can some answer 7 a and b please

pav-90 [236]

Answer:

a.After $15$ second Mr Comer's speed $=$ $1.7 m/s$

b.Distance travelled by Mr.Comer in $15$ seconds $=$ $18.75\ m$

Explanation:

a. Lets recall our first equation of motion $V_{f} =V_{i} + at$

Now we know that $V_{i} = 0.8m/s$ , $a = 0.06\ m/s^2$ and

$t = 15 \ sec$

Plugging the values we have.

$V_{f} =V_{i} + at$

$V_{f} =0.8 + 0.06 \times 15$

$V_{f} =0.8+ 0.9$

$V_{f} =1.7 m/s$

Then Mr.Comer's speed after $15$ sec $=$ $1.7ms^-1$

b.

Lets find the distance and recall our third equation of motion.

$V_{f} ^2-V{i}^2 = 2as$

So $s =$ distance covered.

Dividing both sides with 2a we have.

$\frac{V_{f} ^2-V{i}^2}{2a} = 2as$

Plugging the values.

$\frac{(1.7)^2-(0.8)^2}{2\times0.06} = 2as$

$s= 18.75\ m$

So Mr.Comer will travel a distance of $s= 18.75\ m$ .

4 0

3 years ago

An green hoop with mass mh = 2.8 kg and radius rh = 0.17 m hangs from a string that goes over a blue solid disk pulley with mass

vladimir2022 [97]

The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2

6 0

3 years ago

What’s the power if a student does 2240j of work for 2.8 seconds

Alex787 [66]

Answer:

800 Watts

Explanation:

Power = Work/time

Working in SI units, Power = Watts, Work = Joules, Time = seconds.

Power = 2240J/2.8s = 800 Watts.

7 0

2 years ago

You push your friend, whose mass is 54kg, down a hill so she can go sledding. Her acceleration is 3m/s2. Calculate the amount of

Alecsey [184]

Answer:

18 newtons

Explanation:

Divide weight by speed

3 0

3 years ago

Estimate the final temperature of a mole of gas at 200.0 atm and 19.0°C as it is forced through a porous plug to a final pressur

Yanka [14]

Answer : The final temperature of gas is 266.12 K

Explanation :

According to the Joule-Thomson experiment, it states that when a gas is expanded adiabatically from higher pressure region to lower pressure region, the change in temperature with respect to change in pressure at constant enthalpy is known as Joule-Thomson coefficient.

The formula will be:

$\mu_{J,T}=(\frac{dT}{dP})_H$

or,

$\mu_{J,T}=(\frac{dT}{dP})_H\approx \frac{\Delta T}{\Delta P}$

As per question the formula will be:

$\mu_{J,T}=\frac{T_2-T_1}{P_2-P_1}$ .........(1)

where,

$\mu_{J,T}$ = Joule-Thomson coefficient of the gas = $0.13K/atm$

$T_1$ = initial temperature = $19.0^oC=273+19.0=292.0K$

$T_2$ = final temperature = ?

$P_1$ = initial pressure = 200.0 atm

$P_2$ = final pressure = 0.95 atm

Now put all the given values in the above equation 1, we get:

$0.13K/atm=\frac{T_2-292.0K}{(0.95-200.0)atm}$

$T_2=266.12K$

Therefore, the final temperature of gas is 266.12 K

5 0

3 years ago