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3 years ago

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When you stand up on your feet your feet cover an area of about 0.2 m2. Your feet push on the ground with a pressure of 50 lbs f

or each foot . What is your weight? (Hint: Weight is a force. )
Use the formula Pressure = Force / Area to solve the following problems.

1 answer:

SSSSS [86.1K]3 years ago

7 0

Answer:

135 pounds

Explanation:

and no i didn't copy from the other answer...

maybe i did...

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A 50kg boy runs at a speed of 10.0m/s and jumpsonto a cart

andriy [413]

Answer:

the mass of the cart is 150 kg

Explanation:

given,

mass of boy(m) = 50 kg

speed of boy (v)= 10 m/s

initial velocity of cart (u) = 0

final velocity of cart(V) = 2.5 m/s

mass of the cart(M) = ?

m v + M u = (m + M ) V......................(1)

50× 10 + 0 = (50 + M ) 2.5

M = $\dfrac{500}{2.5} - 50$

M = 150 Kg

hence, the mass of the cart is 150 kg

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3 years ago

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2. Adelia holds a shiny steel spoon with its back (convex surface) facing her eyes at a distance

My name is Ann [436]

Answer:

(a) The convex mirror image, is always upright at all positions, while images formed by concave mirrors are always inverted when the object distance from the mirror is more than the mirrors focal length.

(b) An upright image is not seen for object at a distance from a concave mirror further than the focal length of the mirror, which is the spoon in the question

Therefore, the location of her eyes of approximately, 30 cm, from the mirror is more than the mirror's focal length

Explanation:

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3 years ago

True or false When hydrogen joins with oxygen, a hydrogen bond has been formed.

ladessa [460]

Answer:

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3 years ago

the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force

-Dominant- [34]

Answer:

PART A)

External force will be 75 N

PART B)

distance moved will be 1.125 m

Explanation:

PART A)

Given that net force on the mower is

$F_{net} = 51 N$

now we also know that friction force due to ground is given as

$F_f = 24 N$

now we have

$F_{net} = F_{ext} - F_f$

$51 = F_{ext} - 24$

$F_{ext} = 75 N$

so external force will be 75 N

PART B)

deceleration due to friction when external force is removed from it

$a = \frac{F_f}{m}$

$a = \frac{24}{24} = 1 m/s^2$

now we can find the distance by kinematics

$v_f^2 - v_i^2 = 2 a d$

$0 - 1.5^2 = 2(-1)d$

$d = 1.125 m$

so the distance moved will be 1.125 m

6 0

3 years ago

A close coiled helical spring of round steel wire 10 mm diameter having 10 complete turns with a mean radius of 60 mm is subject

kow [346]

Answer:

The deflection of the spring is 34.56 mm.

Explanation:

Given that,

Diameter = 10 mm

Number of turns = 10

$Radius_{mean} = 60\ mm$

$Diameter_{mean} = 120\ mm$

Load = 200 N

We need to calculate the deflection

Using formula of deflection

$\delta=\dfrac{8pD^3n}{Cd^4}$

Put the value into the formula

$\delta=\dfrac{8\times200\times(120)^3\times10}{80\times10^{3}\times10^4}$

$\delta =34.56\ mm$

Hence, The deflection of the spring is 34.56 mm.

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3 years ago