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3 years ago

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When you are on a huge water slide what forces are there? when will you experience a net force?

1 answer:

Pani-rosa [81]3 years ago

3 0

When you are on a huge water slide, the force present as you slide is the gravitational force. It is because the gravity enables you to slide down the water slide. The net force is the overall forces of the object, so as you slide the water slide, you may experience the net force once you slide down with the gravity and water sliding you down.

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A person is pushing a box.The net external force on the 60-kg box is stated to be 90 N. If the force of friction opposing the mo

Trava [24]

Answer:

a = 1 m/s²

Explanation:

given,

mass of the person = 60 Kg

Net External force exerted = 90 N

force of friction opposing the motion = 30 N

acceleration of the box = ?

Net force = External force applied - frictional force

Net force = 90 - 30

net force = 60 N

we know

F = mass x acceleration

60 = 60 x a

a = 1 m/s²

acceleration of the box is equal to a = 1 m/s²

7 0

3 years ago

First right is brainliest, plz help:)

Aleonysh [2.5K]

They are magnetic

Your welcome;)

5 0

2 years ago

Please help!! giving a lot of points

Readme [11.4K]

Question 1.

mass = 4500 kg
potential energy (p.e) = 67500 J

now, we know :

=》

$p.e = mgh$

=》

$67500 = 4500 \times 10 \times h$

=》

$67500 = 45000 \times h$

=》

$h = \dfrac{67500}{45000}$

=》

$h = 1.5 \: m$

note : if we take acceleration due to gravity as 9.8, then height = 1.53 m

Question 2.

mass = 4500 kg
kinetic energy = 63000 j

we know,

=》

$k.e = \dfrac{1}{2} mv {}^{2}$

=》

$63000 = \dfrac{1}{2} \times 4500 \times {v}^{2}$

=》

${v}^{2} = \dfrac{63000 \times 2}{4500}$

=》

${v}^{2} = 28$

=》

$v = \sqrt{28}$

=》

$v = 2 \sqrt{7} \: \: ms {}^{ - 1}$

or

=》

$5.29 \: \: ms {}^{ - 1}$

7 0

2 years ago

Water waves in a small tank are .06 m long. They pass a given point at a rate of 14.8 waves every three seconds. What is the spe

snow_lady [41]

Answer:

Speed = 0.296m/2

Period = 0.203 s

Explanation:

If by 'long' you mean the wavelength of the waves, then the wavelength $\lambda=0.06m$ .

The frequency $f$ of the waves is 14.8 waves every 3 seconds or

$f=14.8/3 =4.33Hz$ .

Now the relationship between wavelength $\lambda$ , frequency $f$ and speed $v$ of the waves is:

$v=\lambda f$

We put in the values $\lambda=0.06m$ and $f=4.933Hz$ and get:

$\boxed{v=0.06*4.922=0.296m/s}$

Now the period $T$ is just the inverse of the frequency, or

$T=\frac{1}{f}$

$\boxed{T=\frac{1}{4.933}=0.203\:seconds }$

4 0

3 years ago

A 1.11 kg piece of aluminum at 78.3 c is put into a glass with 0.210 kg of water at 15.0

ValentinkaMS [17]

M = mass of aluminium = 1.11 kg

$c_{a}$ = specific heat of aluminium = 900

$T_{ai}$ = initial temperature of aluminium = 78.3 c

m = mass of water = 0.210 kg

$c_{w}$ = specific heat of water = 4186

$T_{wi}$ = initial temperature of water = 15 c

T = final equilibrium temperature = ?

using conservation of heat

Heat lost by aluminium = heat gained by water

M $c_{a}$ ( $T_{ai}$ - T) = m $c_{w}$ (T - $T_{wi}$ )

(1.11) (900) (78.3 - T) = (0.210) (4186) (T - 15)

T = 48.7 c

7 0

3 years ago