All categories

3 years ago

Imagine that person B is more massive than person A in the picture above. 3 points

Physics

1 answer:

mote1985 [20]3 years ago

4 0

Answer:bight

Explanation:

You might be interested in

A 2.00 kg block hangs from a spring balance calibrated in Newtons that is attached to the ceiling of an elevator.(a) What does t

tresset_1 [31]

Answer:

Part a)

$Reading = 2.00 kg$

Part b)

$Reading = 2.00 kg$

Part c)

$Reading = 4.04 kg$

Part d)

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

Explanation:

Part a)

When elevator is ascending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part b)

When elevator is decending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part c)

When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have

$F_{net} = ma$

$T - mg = ma$

$T = mg + ma$

Reading is given as

$Reading = \frac{mg + ma}{g}$

$Reading = 2.00\frac{9.81 + 10}{9.81}$

$Reading = 4.04 kg$

Part d)

Here the speed of the elevator is constant initially

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

3 0

3 years ago

Help me out with this pe question plz :c

blagie [28]

Answer:

the first one is a group 1 and the second one is d all of the above

Explanation:

6 0

3 years ago

Read 2 more answers

A 50.0 N box sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at on

Vikki [24]

Answer:

-4.0 N

Explanation:

Since the force of friction is the only force acting on the box, according to Newton's second law its magnitude must be equal to the product between mass (m) and acceleration (a):

$F_f = ma$ (1)

We can find the mass of the box from its weight: in fact, since the weight is W = 50.0 N, its mass will be

$m=\frac{W}{g}=\frac{50.0 N}{9.8 m/s^2}=5.1 kg$

And we can fidn the acceleration by using the formula:

$a=\frac{v-u}{t}$

where

v = 0 is the final velocity

u = 1.75 m/s is the initial velocity

t = 2.25 s is the time the box needs to stop

Substituting, we find

$a=\frac{0-1.75 m/s}{2.25 s}=-0.78 m/s^2$

(the acceleration is negative since it is opposite to the motion, so it is a deceleration)

Therefore, substituting into eq.(1) we find the force of friction:

$F_f = (5.1 kg)(-0.78 m/s^2)=-4.0 N$

Where the negative sign means the direction of the force is opposite to the motion of the box.

6 0

3 years ago

A square plate of side 9 m is submerged in water at an incline of 60∘ with the horizontal. Its top edge is located at the surfac

Tpy6a [65]

Answer:

The force on one side of the plate is 3093529.3 N.

Explanation:

Given that,

Side of square plate = 9 m

Angle = 60°

Water weight density = 9800 N/m³

Length of small strip is

$y=\dfrac{\Delta y}{\sin60}$

$y=\dfrac{2\Delta y}{\sqrt{3}}$

The area of strip is

$dA=\dfrac{9\times2\Delta y}{\sqrt{3}}$

We need to calculate the force on one side of the plate

Using formula of pressure

$P=\dfrac{dF}{dA}$

$dF=P\times dA$

On integrating

$\int{dF}=\int_{0}^{9\sin60}{\rho g\times y\times 6\sqrt{3}dy}$

$F=9800\times6\sqrt{3}(\dfrac{y^2}{2})_{0}^{9\sin60}$

$F=9800\times6\sqrt{3}\times(\dfrac{(9\sin60)^2}{2})$

$F=3093529.3\ N$

Hence, The force on one side of the plate is 3093529.3 N.

7 0

4 years ago

You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene

uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is $150\,^{\circ}C$ . According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

$\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T)$ (1)

Where:

$\rho_{k}$ - Density of kerosene, measured in kilograms per cubic meter.

$V_{k}$ - Volume of kerosene, measured in cubic meters.

$c_{k}$ , $c_{t}$ - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

$T_{k,o}$ , $T_{t,o}$ - Initial temperatures of kerosene and tin, measured in degrees Celsius.

$T$ - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

$V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}$

If we know that $m_{t} = 1.83\,kg$ , $c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{t,o} = 88\,^{\circ}C$ , $T_{k,o} = 24.0\,^{\circ}C$ , $T = 57\,^{\circ}C$ , $c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C}$ and $\rho_{k} = 820\,\frac{kg}{m^{3}}$ , then the volume of the liquid needed to accomplish this task without boiling is:

$V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}$

$V_{k} = 2.273\times 10^{-4}\,m^{3}$

$V_{k} = 0.273\,L$

0.273 liters are needed to accomplish this task without boiling.

3 0

3 years ago