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3 years ago

8

If a total force exerted by water in a container with a bottom area of 2 square meters is 450 newtons, what's the water pressure

at the bottom of the container? A. 0.900 kPa B. 0.225 kPa C. 0.575 kPa D. 0.300 kPa

2 answers:

m_a_m_a [10]3 years ago

7 0

its B 0.225kPa using the formula p=f/A then change the pascals into kpa

Ipatiy [6.2K]3 years ago

7 0

The Correct answer to this question for Penn Foster Students is: 0.225 kPa

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A stunt man jumps from the top of a building and lands 10 meters below his initial height. In case A, the stunt man lands on a s

krek1111 [17]

Answer:

Explanation:

The stunt will likely sustain serious injury in case of concrete blocks because the average force acting on the person will be more because concrete blocks do not squeeze to provide more time for the force to act on the body instead it acts for a small amount of interval.

$Impulse=F_{avg}\times \Delta T$

As impulse is constant so time requires to act force on the body is more as compared to concrete block and thus average force in mattress case is less.

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3 years ago

what will the stopping distance be a a 3000-kg car if -3000N of force are applied when the car is traveling 10 m/s

Inga [223]

Answer:

50 m

Explanation:

Acceleration= force/mass

3000/3000=1m/s^-2

Applying equation of motion:

V^2=U^2+2as; V is final velocity, u is initial velocity, a is acceleration and s is the distance covered.

0=10^2 -2*1s;

Solve for s

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3 years ago

The force of magnitude F acts along the edge of the triangular plate. Determine the moment of F about point O. Find the general

Sever21 [200]

Answer:

The moment is 81.102 k N-m in clockwise.

Explanation:

Given that,

Force = 260 N

Side = 580 mm

Distance h = 370 mm

According to figure,

Position of each point

$O=(0,0)$

$A=(0,-b)$

$B=(h,0)$

We need to calculate the position vector of AB

$\bar{AB}=(h-0)i+(0-(-b))j$

$\bar{AB}=hi+bj$

We need to calculate the unit vector along AB

$u_{AB}=\dfrac{\bar{AB}}{|\bar{AB}|}$

$u_{AB}=\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}}$

We need to calculate the force acting along the edge

$\hat{F}=F(u_{AB})$

$\hat{F}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})$

We need to calculate the net moment

$\hat{M}=\hat{F}\times OA$

Put the value into the formula

$\hat{M}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})\times(-b\hat{j})$

$\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}((h\hat{i}+b\hat{j})\times(-b\hat{j}))$

$\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}(-bh\hat{k})$

$\hat{M}=-\dfrac{bhF}{\sqrt{h^2+b^2}}$

Put the value into the formula

$\hat{M}=-\dfrac{580\times10^{-3}\times370\times10^{-3}\times260}{\sqrt{(370\times10^{-3})^2+(580\times10^{-3})^2}}$

$\hat{M}=-81.102\ \hat{k}\ N-m$

Negative sign shows the moment is in clockwise.

Hence, The moment is 81.102 k N-m in clockwise.

5 0

3 years ago

A ball is dropped from the top of a cliff. By the time it reaches the ground, all of its gravitational potential energy has been

Katena32 [7]

Answer:

20 m

Explanation:

Initial potential energy = final kinetic energy

mgh = 1/2 mv²

gh = 1/2 v²

h = v² / (2g)

Given v = 20 m/s and g = 10 m/s²:

h = (20 m/s)² / (2 × 10 m/s²)

h = 20 m

4 0

3 years ago

A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 4.0 m/s, skat

Cerrena [4.2K]

Answer:

a) $t=59.817652833125294s \approx 59.8s$

b) $D_1=894.01m$

Explanation:

From the question we are told that

Speed of opposing player $V_2=4.0m/s$

First player chase his opponent after $t=2.80t$

Acceleration of first player $a=0.14 m/s2$

Let time of catch be $t_c$

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$D_1=\frac{1}{2}(0.14)t^2$

$D_1=0.07t^2$

Distance to covered Second opponent

$D_2=(2.8+t)*4$

Generally when first opponent catch the second opponent it is represented mathematically as

$D_2=D_1$

$0.07t^2=(2.8+t)*4$

$0.07t^2=11.2+4t$

$0.07t^2-4t-11.2$

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b)Generally the the total time traveled by the first opponent is mathematically given as

$D_1=\frac{t^2}{4}$

$D_1=\frac{59.8^2}{4}$

$D_1=894.01m$

3 0

3 years ago