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3 years ago

Doe anyone get this

Physics

1 answer:

8_murik_8 [283]3 years ago

4 0

Answer:

we know a = F/ M

Explanation:

2 m/s²
0.19 m/s²
9.25 m/ s²
0.04 m/s²
100.39 m/s²

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A nonconducting container filled with 25 kg of water at 23°C is fitted with a stirrer, which is made to turn by gravity acting o

Paul [167]

Explanation:

Given that,

Weight of water = 25 kg

Temperature = 23°C

Weight of mass = 32 kg

Distance = 5 m

(a). We need to calculate the amount of work done on the water

Using formula of work done

$W=mgh$

$W=32\times9.8\times5$

$W=1568\ J$

The amount of work done on the water is 1568 J.

(b). We need to calculate the internal-energy change of the water

Using formula of internal energy

The change in internal energy of the water equal to the amount of the work done on the water.

$\Delta U=W$

$\Delta U=1568\ J$

The change in internal energy is 1568 J.

(c). We need to calculate the final temperature of the water

Using formula of the change internal energy

$\Delta U=mc_{p}\Delta T$

$\Delta U=mc_{p}(T_{2}-T_{1})$

$T_{2}=T_{1}+\dfrac{\Delta U}{mc_{p}}$

$T_{2}=23+\dfrac{1568}{25\times4.18\times10^{3}}$

$T_{2}=23.01^{\circ}\ C$

The final temperature of the water is 23.01°C.

(d). The amount of heat removed from the water to return it to it initial temperature is the change in internal energy.

The amount of heat is 1568 J.

Hence, This is the required solution.

6 0

3 years ago

Determine the amount of work done when a crane lifts a 100-N block form 2m above the ground to 6m above the ground

Dima020 [189]

Data given:

F=100N

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A=?

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A=F×s

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3 years ago

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3 years ago

Read 2 more answers

A motor is connected to the smaller 0.2 m radius pulley and cable, lifting 150 kg mass. Neglecting the mass/weight of the cable

-Dominant- [34]

Answer:

324.3Nm

Explanation:

The torque is given by the equation

$\tau=r\ X\ F$

in this case the vectors r and F are perpendicular between them, thus:

$\tau=rF$

The forces acting on the mass are:

$T-Mg=Ma$ (1)

where T is the tension of the cable, M is the mass and a is the acceleration.

Furthermore, we have that the acceleration is:

$a=\frac{v-v_0}{t}=\frac{10m/s-0m/s}{10s}=1\frac{m}{s^2}$

By replacing in (1) we can obtain:

$T=Ma-Mg=M(a-g)=(150kg)(1\frac{m}{s^2}+9.8\fac{m}{s^2})=1620N$

The force T produces the torque on the pulley, hence:

$\tau=rT=(0.2m)(1620N)=324Nm$

the answer is 324.4Nm

hope this helps!

7 0

3 years ago

Doe anyone get this ​

Doe anyone get this