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Rus_ich [418]
3 years ago
5

Bob is pushing a box across the floor at a constant speed of 1.4 m/s, applying a horizontal force whose magnitude is 65 N. Alice

is pushing an identical box across the floor at a constant speed of 2.8 m/s, applying a horizontal force. (a) What is the magnitude of the force that Alice is applying to the box? F = N (b) With the two boxes starting from rest, explain qualitatively what Alice and Bob did to get their boxes moving at different constant speeds. In order to keep the box moving twice as fast, Alice had to apply a constant force that was twice as large as the force that Bob applied. Each initially applied a force bigger than static friction to get the box moving and accelerating, then when the desired final speed was achieved they reduced the force to make the net force zero.
Physics
1 answer:
bija089 [108]3 years ago
5 0

Answer:

a) 65 N

b) Each initially applied a force bigger than static friction to get the box moving and accelerating, then when the desired final speed was achieved they reduced the force to make the net force zero.

Explanation:

a) Here, both boxes are identical. They have different velocities but the velocities are constant hence the force Alice is applying is 65 N. Constant velocity means there is no acceleration.

b) The acceleration that Alice provided must have been different than that of Bob. But after reaching their desired speed they stopped accelerating. This would make the net force zero.

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(d) If η = 40% and TH = 427°C, what is TC, in °C?
Brrunno [24]

Answer:

T_C=256.2^{\circ}C

Explanation:

Given that,

Efficiency of heat engine, \eta=40\%=0.4

Temperature of hot source, T_H=427^{\circ}C

We need to find the temperature of cold sink i.e. T_C. The efficiency of heat engine is given by :

\eta=1-\dfrac{T_C}{T_H}

T_C=(1-\eta)T_H

T_C=(1-0.4)\times 427

T_C=256.2^{\circ}C

So, the temperature of the cold sink is 256.2°C. Hence, this is the required solution.

3 0
3 years ago
A tiger is running across a field. What would happen if the hydrogen ion pumps into the tigers mitochondria stopped functioning
galben [10]

Answer:

The tiger would not be able to produce glucose causing it to stop running

Explanation:

Since the mitochondria is in charge of producing ATP the tiger would not be able to use any glucose causing it to not be able to run.

3 0
3 years ago
The free-body diagram of a crate is shown. A free body diagram with 4 force vectors. The first vector is pointing downward, labe
satela [25.4K]

The net force acting on the crate is determined as 176 N to the left.

<h3>Net force acting on the crate</h3>

The net force acting on the crate is calculated as follows;

∑F = F1 + F2 + F3 + F4

F(net) = -440y + 176x + 440y - 352x

F(net) = -176 x

The resultant force is pointing in negative x direction.

Thus, the net force acting on the crate is determined as 176 N to the left.

Learn more about net force here: brainly.com/question/14361879

#SPJ1

8 0
2 years ago
8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s
allsm [11]

Answer:

a) \omega = 50\,\frac{rad}{s}, b) \omega = 0\,\frac{rad}{s}

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

\tau = F \cdot r

\tau = m\cdot a \cdot r

\tau = m \cdot \alpha \cdot r^{2}

Where \alpha is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}

\alpha = -3\,\frac{rad}{s^{2}}

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)

\omega = 50\,\frac{rad}{s}

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}

t = 66.667\,s

Since t > 66.667\,s, then the angular velocity is equal to zero. Therefore:

\omega = 0\,\frac{rad}{s}

7 0
3 years ago
An ice chest at a beach party contains 12 cans of soda at 4.05 °C. Each can of soda has a mass of 0.35 kg and a specific heat ca
BARSIC [14]
Below is the solution:

Heat soda=heat melon 
<span>m1*cp1*(t-t1)=m2*cp2*(t2-t); cp2=cpwater </span>
<span>12*0.35*3800*(t-5)=6.5*4200*(27-t) </span>
<span>15960(t-5)=27300(27-t) </span>
<span>15960t-136500=737100-27300t </span>
<span>43260t=873600 </span>
<span>t=873600/43260 </span>
<span>t=20.19 deg celcius</span>
4 0
3 years ago
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