Which of the following is an example of a buoyant force acting on an object?

How much does a dog of 10kg mass weigh in newtons?

ser-zykov [4K]

Answer:

98.0665 newtons

have a good day :]

7 0

3 years ago

a student is pushing a 50 kilogram cart with a force of 500 newtons another students measures the speed of the cart and finds th

Irina-Kira [14]

The force of friction is 300 N

Explanation:

We can solve the problem by applying Newton's second law of motion: in fact, the net force acting on an object is equal to the product between the mass of the object and its acceleration. So we can write

$\sum F = ma$

where

$\sum F$ is the net force acting on the object

m is its mass

a is its acceleration

For the cart in this problem, we have two forces acting on it:

- The force of push, F = 500 N, forward

- The force of friction, $F_f$ , backward

So Newton's second law can be rewritten as

$F-F_f = ma$

where

m = 50 kg

$a=4 m/s^2$ is the acceleration of the cart

And solving for $F_f$ , we find the force of friction:

$F_f = F-ma=500-(50)(4)=300 N$

Learn more about force of friction:

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4 0

3 years ago

A 24-cm-diameter vertical cylinder is sealed at the top by a frictionless 15 kg piston. The piston is 90 cm above the bottom whe

abruzzese [7]

Answer:

(A) $P_i=3249.41\ Pa$

(B) $V_f=0.3358\ m$

Explanation:

Given:

diameter of the cylinder, $d= 0.24\ m$

mass of piston sealing on the top, $m_p=15\ kg$

initial temperature of the piston, $T_i=315+273= 588\ K$

initial height of piston, $h_i=0.9\ m$

atmospheric pressure on the piston, $p_a=1 atm=101325\ Pa$

(A)

Initial pressure of gas is the pressure balanced by the weight of piston:

$P_i=\frac{m_p.g}{\pi.d^2\div 4}$

$P_i=\frac{15\times 9.8}{\pi\times (0.24^2\div 4)}$

$P_i=3249.41\ Pa$

Which is gauge pressure because it is measured with respect to the atmospheric pressure.

(B)

Given:

Final temperature, $T_f=18+273=291\ K$

Now, volume of air initially in the cylinder:

$V_i=\pi.d.h_i$

$V_i=\pi\times 0.24\times 0.9$

$V_i=0.6786\ m^3$

Using gas law:

$\frac{P_i V_i}{T_i}= \frac{P_f V_f}{T_f}$ ........................................(1)

∵In every condition of equilibrium the gas pressure will be balanced by the weight of the piston so it is an isobaric transition.

∴ $P_i=P_f$

Hence eq. (1) is reduced to:

$\frac{V_i}{T_i}= \frac{V_f}{T_f}$

putting respective values:

$\frac{0.6786}{588}= \frac{V_f}{291}$

$V_f=0.3358\ m$

6 0

4 years ago

A ball with a mass of 3.1 kg is moving in a uniform circular motion upon a horizontal surface. The ball is attached at the cente

inessss [21]

Answer:

3.46 seconds

Explanation:

Since the ball is moving in circular motion thus centripetal force will be acting there along the rope.

The equation for the centripetal force is as follows -

$F=\frac{mv^2}{r}$

Where, $m$ is the mass of the ball, $v$ is the speed and $r$ is the radius of the circular path which will be equal to the length of the rope.

This centripetal force will be equal to the tension in the string and thus we can write,

$20.4 = \frac{3.1\times v^2}{2}$

and, $v^2 = 13.16$

Thus, $v = 3.63$ m/s.

Now, the total length of circular path = circumference of the circle

Thus, total path length = 2πr = 2 × 3.14 × 2 = 12.56 m

Time taken to complete one revolution = $\frac{\text {Path length} }{\text {Speed}}$ = $\frac{12.56}{3.63}$ = 3.46 seconds.

Thus, the mass will complete one revolution in 3.46 seconds.

4 0

3 years ago

Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diamete

Sonbull [250]

Answer:

work done in pumping the entire fuel is 466587 J

Explanation:

weight of the gasoline per volume = 6600 N/m^3

diameter of the tank = 3 m

length of the tank = 2 m

height of the tractor tank above the top of the tank = 5 m

work done in pumping fuel to this height = ?

First, we find the volume of the fuel

since the tank is cylindrical, we assume that the fuel within also takes the cylindrical shape.

Also, we assume that the fuel completely fills the tank.

volume of a cylinder = $\pi r^{2}l$

where r = radius = diameter ÷ 2 = 3/2 = 1.5 m

volume of the cylinder = 3.142 x $1.5^{2}$ x 2 = 14.139 m^3

we then find the total weight of the fuel in Newton

total weight = (weight per volume) x volume

total weight = 6600 x 14.139 = 93317.4 N

work done = (total weight of the fuel) x (height through which the fuel is pumped)

work done in pumping = 93317.4 x 5 = 466587 J

8 0

3 years ago