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pantera1 [17]
3 years ago
7

A block is being pulled by a rope in a straight line across a horizontal floor which is non frictionless. The rope is pulling on

the block with a force of 60N, and the frictional force exerted on the block by the floor is 20N. What is the force with which the block is pulling back on the rope?
Physics
1 answer:
ikadub [295]3 years ago
6 0

Answer:

60 N

Explanation:

We can answer the question by using Newton's third law, which states that:

"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"

In this situation, we can identify the rope as object A, and the block as object B.

We are told that the rope exerts a force of 60 N on the block: if we apply Newton's third law, therefore, we can say that the block will also exert an equal (60 N) and opposite (in direction) force on the rope.

You might be interested in
A child pulls on a wagon with a force of 75 N. If the wagon moves a total of 42 m in 3.1 min, what is the average power delivere
exis [7]

Answer:

16.96 W

Explanation:

Power: This can be defined as the rate at which work is done by an object. The S.I unit of power is Watt(W).

From the question,

P = (F×d)/t....................... Equation 1

Where P = power, F = force, d = distance, t = time.

Given: F = 75 N, d = 42 m, t = 3.1 min = 3.1×60 = 186 s

Substitute these values into equation 1

P = (75×42)/186

P = 16.94 W

Hence the average power delivered by the child  = 16.96 W

4 0
3 years ago
2(a)
Tamiku [17]

Answer:

2a.

a=1.13ms^-2

2b.

S=277m

2c.

V=27.7ms-¹

Explanation:

Initial Velocity (U)=22m/s¹

Final Velocity (V)=43m/s²

Time(t) =18.6s

a. a=V-U/t

a=43-22/18.6

a=1.129

a=1.13m/s²

2b.

S=ut+1/2 at²

s=22(10)+1/2×1.13(10)²

s= 220+0.57(10)²

s= 220+0.57(100)

s= 220+57

s=277m

2c.

V=U+AT

V=22+1.13(5)

V=22+5.65

V=22+5.7

V=27.7m/s¹

6 0
3 years ago
Water pollution that elevates the temperature of the water​
juin [17]

Answer:

An increase in the air temperature will cause water temperatures to increase as well. As water temperatures increase, water pollution problems will increase, and many aquatic habitats will be negatively affected.

Explanation:

Lower levels of dissolved oxygen due to the inverse relationship that exists between dissolved oxygen and temperature. As the temperature of the water increases, dissolved oxygen levels decrease.

Increases in pathogens, nutrients and invasive species.

Increases in concentrations of some pollutants such as ammonia and pentachlorophenol due to their chemical response to warmer temperatures.

Increase in algal blooms (Photo of algal blooms).

Loss of aquatic species whose survival and breeding are temperature dependent.

Change in the abundance and spatial distribution of coastal and marine species and decline in populations of some species.

Increased rates of evapotranspiration from waterbodies, resulting in shrinking of some waterbodies such as the Great Lakes.

7 0
4 years ago
Water is flowing into a factory in a horizontal pipe with a radius of 0.0183 m at ground level. This pipe is then connected to a
timama [110]

Answer:

0.0168 m^3/s

Explanation:

We are given that

r_1=0.0183 m

h_1=0

r_2=0.0420 m

h_2=12.6 m

Let P_1=P_2=P

By using Bernoulli theorem

P+\frac{1}{2}\rho v^2_1+\rho gh_1=P+\frac{1}{2}\rho v^2_2+\rho gh_2

\frac{1}{2}\rho v^2_1+\rho gh_1=\frac{1}{2}\rho v^2_2+\rho gh_2

v^2_1+2gh_1=v^2_2+2gh_2

A_1v_1=A_2v_2

v_1=\frac{A_2v_2}{A_1}

(\frac{A_2}{A_1})^2v^2_2+2g\times 0=v^2_2+2\times 9.8\times 12.6

(\frac{\pi r^2_2}{\pi r^2_1})^2v^2_2-v^2_2=246.96

v^2_2((\frac{r^2_2}{r^2_1})^2-1)=246.96

v^2_2=246.96\frac{r^4_1}{r^2_4-r^4_1}

v_2=\sqrt{246.96\frac{r^4_1}{r^4_2-r^4_1}}

v_2=\sqrt{246.96\times \frac{(0.0183)^4}{(0.042)^4-(0.0183)^4}}

v_2=3.038 m/s

Volume flow rate =A_2v_2

Volume flow rate =\pi r^2_2v_2=\pi (0.042)^2\times 3.038=0.0168 m^3/s

3 0
3 years ago
5. Aunt Jane weighs 145 pounds. What is her weight in Newtons?
Alex_Xolod [135]

Answer:

i think its 644

Explanation:

5 0
3 years ago
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