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2 years ago

A box has pushed across the floor for 5.2 M with a 48 N force how much work is done?

1 answer:

8 0

Work is force x displacement
So force is 48 and displacement is 5.2 as given

Therefore 48 x 5.2
Which is 249.6
Approximately 250 J (joules)

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A wooden cylinder of length L and cross-sectional area A is partially submerged in a liquid with the axis of the cylinder orient

SCORPION-xisa [38]

Answer:

$F=\rho_LAdg$

Explanation:

The buoyant force F is equal to the weight of the displaced fluid. The weight of the displaced fluid is $W=m_dg$ , where $m_d$ is the mass of the displaced fluid. The mass of the displaced fluid is $m_d=\rho_LV_d$ , where $\rho_L$ is the density of the fluid and $V_d$ is the displaced volume, which is equal to the submerged volume of the cilinder $V_d=V_s=Ad$ .

Putting all together we have:

$F=W=m_dg=\rho_LV_dg=\rho_LAdg$

7 0

3 years ago

On my bike I was able to travel 40 miles in one hour from this information we can determine the bikes

valentinak56 [21]

You have a distance and a time so you can work out the bikes speed which would be distance/time so 40mph.

4 0

2 years ago

A race car travels 40 m/s around a banked (45° with the horizontal) circular (radius = 0.20 km) track. What is the magnitude of

OLEGan [10]

Answer:

c) $F_{net} = 0.640 kN$

Explanation:

As we know that resultant force is the net force that is acting on the system

As per Newton's II law we know that net force is product of mass and acceleration

so we will have

$F_{net} = ma$

here we know

m = 80 kg

for circular motion acceleration is given as

$a_c = \frac{v^2}{R}$

$a_c = \frac{40^2}{200} = 8 m/s^2$

now we have

$F_{net} = 80 \times 8$

$F_{net} = 640 N$

$F_{net} = 0.640 kN$

7 0

3 years ago

Which graph shows an object that is dropped?

Olegator [25]

Answer:

I'm pretty sure it's the third one where velocity goes from positive to negative

Explanation:

the positive velocity is before the object hits the ground and the negative is after

8 0

3 years ago

A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein

RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of $8m^{3}/s$ . As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi r^{2}h$

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

$\frac {r}{h}=\frac{4}{5}$