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emmasim [6.3K]
2 years ago
9

Anna needs to move a box of paperback books across the room. If she applies a force of 20 newtons to the box, what is the magnit

ude of the force exerted by the box on Anna?
Physics
2 answers:
vlabodo [156]2 years ago
5 0
Newton's third law tells us that for every force there is an equal and opposite force.  This means that if Anna exerts a force of 20 Newtons on the box, the box exerts a force of 20 Newtons on Anna.
ycow [4]2 years ago
3 0

Answer:

The magnitude of the force exerted by the box on Anna is 20 newtons.

Explanation:

It is given that, Anna needs to move a box of paperback books across the room. If she applies a force of 20 newtons to the box. We need to find the magnitude of the force exerted by the box on Anna.

The force exerted by the box on Anna is 20 newtons but it acts in the opposite direction as per Newton's third law of motion. The third law of motion explains the impact of forces i.e. action and reaction force.

So, the magnitude of the force exerted by the box on Anna is also 20 newtons. Hence, this is the required solution.

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Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit
snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is  

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then  

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity=  (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

is

Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

Inlet volume = inlet velocity*area of circle=\pi  r^{2}*inlet velocity

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

4 0
2 years ago
At a local swimming pool, the diving board is elevated h = 5.5 m above the pool's surface and overhangs the pool edge by L = 2 m
Margaret [11]

Answer:

Part a)

t = \sqrt{\frac{2h}{g}}

Part b)

t = 1.06 s

Part c)

L  = 4.86 m

Explanation:

Part a)

The height of the diving board is given as

h = 5.5 m

now the speed of the diver is given as

v_0 = 2.7 m/s

when the diver will jump into the water then his displacement in vertical direction is same as that of height of diving board

So we will have

y = v_y t + \frac{1}{2}at^2

h = 0 + \frac{1}{2}gt^2

t = \sqrt{\frac{2h}{g}}

Part b)

t = \sqrt{\frac{2h}{g}}

plug in the values in the above equation

t = \sqrt{\frac{2(5.5 m)}{9.81}

t = 1.06 s

Part c)

Horizontal distance moved by the diver is given as

d = v_0 t

d = 2.7 \times 1.06

d = 2.86 m

so the distance from the edge of the pool is given as

L = 2.86 + 2

L  = 4.86 m

4 0
2 years ago
Separating the electron from the proton in a hydrogen atom takes 2.18 ✕ 10−18 j of work. through what electric potential differe
My name is Ann [436]
Electric potential = work done/charge of electron = 2.18×10⁻¹⁸/1.6×10⁻¹⁹
                                                                              =  13.625 V
6 0
1 year ago
What sentence best supports the statement that hormones are involved in the regulation of homeostasis? A. The hormone cortisol s
juin [17]

B. The hormone erythropoeitin increases the production of red blood cells when oxygen levels are low.

3 0
2 years ago
Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for
g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0
2 years ago
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