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Ray Of Light [21]

3 years ago

11

A rectangular block weighs 240 N. the area of the block in contact with the floor is 20 cm2.calculate the pressure on the floor(

give your answer in N/cm2
Help me Fast as you can,,

1 answer:

maks197457 [2]3 years ago

7 0

Answer:

12 N/cm²

Explanation:

From the question given above, the following data were obtained:

Weight (W) of block = 240 N

Area (A) = 20 cm²

Pressure (P) =?

Next, we shall determine the force exerted by the block. This can be obtained as follow:

Weight (W) of block = 240 N

Force (F) =.?

Weight and force has the same unit of measurement. Thus, we force applied is equivalent to the weight of the block. Thus,

Force (F) = Weight (W) of block = 240 N

Force (F) = 240 N

Finally, we shall determine the pressure on the floor as follow:

Force (F) = 240 N

Area (A) = 20 cm²

Pressure (P) =?

P = F/A

P = 240 / 20

P = 12 N/cm²

Therefore, the pressure on the floor is 12 N/cm².

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A punter drops a 2.0 kg ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at

pav-90 [236]

Answer:

Explanation:

Given

mass of drop $m=2 kg$

height of fall $h=1 m$

ball leaves the foot with a speed of 18 m/s at an angle of $55^{\circ}$

Velocity of ball just before the collision with the floor

$u^=2gh$

$u=\sqrt{2gh}$

$u=\sqrt{2\times 9.8\times 1}=4.42 m/s$

Impulse delivered in Y direction

$J_y=m(v\sin (55)-(-u))$

$J_y=2(18\sin (55)+4.42)$

$J_y=38.32 kg-m/s$

Impulse in x direction

$J_x=m\times v\cos (55)$

$J_x=2\times 4.42\cos (55)=20.646$

$J_{net}=\sqrt{J_x^2+J_y^2}$

$J_{net}=\sqrt{(38.32)^2+(20.64)^2}$

$J_{net}=43.52 kg-m/s$

at an angle of $\tan \phi =\frac{J_y}{J_x}=\frac{38.32}{20.64}$

$\phi =tan^{-1}(1.856)$

$\phi =61.7^{\circ}$

7 0

3 years ago

Two ice skaters, Daniel (mass 70.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while a

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Answer:

a) $v=7.32m/s$

b) $\alpha =-35$ º

c) $ΔK=-1094.62J$

Explanation:

From the exercise we know that the collision between Daniel and Rebecca is elastic which means they do not stick together

So, If we analyze the collision we got

$p_{1x}=p_{2x}$

To simplify the problem, lets name D for Daniel and R for Rebecca

a) $p_{D1x}+p_{R1x}=p_{D2x}+p_{R2x}$

Since Daniel's initial velocity is 0

$m_{R}v_{Rx}=m_{D}v_{D2x}+m_{R}v_{R2x}$

$v_{D2x}=\frac{m_{R}*v_{R1x}-m_{R}*v_{R2x}}{m_{D}}$

$v_{D2x}=\frac{(45kg)(14m/s)-(45kg)(8cos(55.1)m/s)}{(70kg)}=6m/s$

Now, lets analyze the movement in the vertical direction

$p_{1y}=p_{2y}$

Since $p_{1y}=0$

$0=m_{D}v_{D2y}+m_{R}v_{R2y}$

$v_{D2y}=-\frac{m_{R}v_{2Ry}}{m_{D}}=-\frac{(45kg)(8sin(55.1)m/s)}{(70kg)}=-4.21m/s$

Now, we can find the magnitude of Daniel's velocity after de collision

$v_{D}=\sqrt{(6m/s)^2+(-4.21m/s)^2}=7.32m/s$

b) To know whats the direction of Daniel's velocity we need to solve the arctan of the angle

$\alpha =tan^{-1}(\frac{v_{y}}{v_{x}})=tan^{-1}(\frac{-4.21}{6})=-35$ º

c) The change in the total kinetic energy is:

ΔK= $K_{2}-K_{1}$

ΔK= $\frac{1}{2}[(45kg)(8m/s)^2+(70kg)(7.32m/s)^2-(45kg)(14m/s)^2]=-1094.62J$

That means that the kinetic energy decreases

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3 years ago

PLEASE HELP!!! THANKS I GIVE BRAINLIEST !!A student examines the effect of the number of D batteries in a closed circuit on the

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Answer:

The batteries

Explanation:

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3 years ago

A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1

nikklg [1K]

Answer:

Part a)

$F_n = 306 N$

Part B)

$v = 12.1 m/s$

Explanation:

Part A)

At the top of the hump the force on the rider is

1) Normal force

2) weight

so here we know that

$mg - F_n = \frac{mv^2}{R}$

$F_n = mg - \frac{mv^2}{R}$

$F_n = (100)(9.81) - \frac{100(9^2)}{12}$

$F_n = 306 N$

Part B)

At the top of the loop we will have

$F_n + mg = \frac{mv^2}{R}$

in order to remain in contact the normal force must be just greater than zero

so we will have

$mg = \frac{mv^2}{R}$

$v = \sqrt{Rg}$

$v = \sqrt{15\times 9.81}$

$v = 12.1 m/s$

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3 years ago

A missile is launched upward with a speed that is half the escape speed. What height (in radii of Earth) will it reach?

timofeeve [1]

Answer:

The height (h) will be: $\frac{3}{4}R =h$

Explanation:

The scape speed equation is given by:

$v_{scape}=\sqrt{\frac{2GM}{R}}$

Now, the speed of the missile is

$v_{missile}=\frac{1}{2}v_{scape}$

$v_{scape}=\frac{1}{2}\sqrt{\frac{2GM}{R}}$

Using the conservation of energy, we can find the maximu height of the missile.

$E_{i}=E_{f}$

$\frac{1}{2}mv_{scape}^{2}-mgR =-mgh$

$\frac{1}{2}\frac{2GM}{4R}-gR =-gh$

$\frac{GM}{4R}-gR =gh$

Let's recall that g = GM/R², using the equivalence principle. When R is the radius of the earth and M is the mass of the earth.

$\frac{1}{4}gR-gR =-gh$

$\frac{1}{4}R-R =-h$

Therefore the height (h) will be:

$\frac{3}{4}R =h$

I hope it helps you!

4 0

3 years ago