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3 years ago

a student pushes a 40-N block across the floor for a distance 0f 10 m how much work was done to move the block?

Physics

2 answers:

slava [35]3 years ago

6 0

W=f*d
= 40N * 10m
work= 400 joules

bearhunter [10]3 years ago

3 0

We don't know how much force s/he had to use to push the block.
If the floor was smooth and wet, it needed very little force.
If there was sandpaper or stones on the floor, it took a lot of force.

Whatever amount of force s/he needed to use, the work done was

Work = (amount of force, in Newtons) x (10 m)

The unit is Joules.

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Two charged objects are 1 meter apart. Calculate the magnitude of the electric force between them if the two charges are +1.0 μC

Solnce55 [7]

Answer:

0.00899 N

Explanation:

The magnitude of the electrostatic force between two charges is given by the equation:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the charges

r is the distance between the two charges

And the force is:

- Repulsive if the two charges have same sign

- Attractive if the two charges have opposite sign

In this problem we have:

$q_1=1.0\mu C = 1.0\cdot 10^{6}C$ (charge of object 1)

$q_2=1.0\mu C = 1.0\cdot 10^{6}C$ (charge of object 2)

r = 1 m (separation between the objects)

So, the electric force is

$F=(8.99\cdot 10^9)\frac{(1.0\cdot 10^{-6})^2}{1^2}=0.00899 N$

5 0

4 years ago

A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal

AysviL [449]

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

$v=\sqrt{v_x^2+v_y^2}$ (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

$v_y^2=v_{oy}^2+2gh$ (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

$v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}$

vx is calculated by using the information about the horizontal range of the ball:

$R=v_o\sqrt{\frac{2h}{g}}$ (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

$v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}$

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

$v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}$

The velocity of the ball just before it touches the ground is 46.99 m/s

5 0

4 years ago

a 8000kg tow truck pulls a 3000kg car horizontally into a garage. if the motor of the tow truck exerts a forward thrust on the t

Evgen [1.6K]

Equation 1 :
m1 : m1a =T
Equation 2:m2 : m2a= F - T

Adding 1 and 2a=0.073

Placing in equation 1

we get T = 218.18N

8 0

3 years ago

The tip of the second hand of a clock moves in a circle of 20 cm circumference. In one minute the hand makes a complete revoluti

Cerrena [4.2K]

Answer:

$v_{avg} = 0$

Explanation:

As we know that average velocity is defined as the ratio of total displacement of the object and its time interval.

so here we can say

$v_{avg} = \frac{displacement}{time}$

now we know that in one complete revolution the total displacement of the tip of the seconds hand is zero

because it will have same position after one complete revolution from where it starts

so here we can say that the average velocity will be zero

$v_{avg} = 0$

7 0

3 years ago

An electric furnace is to melt 40 kg of aluminium/hour. The initial temperature of aluminium is 32°C. Given that aluminium has s

gizmo_the_mogwai [7]

Answer:

Part a)

P = 13.93 kW

Part b)

R = 8357.6 Cents

Explanation:

Part A)

heat required to melt the aluminium is given by

$Q = ms\Delta T + mL$

here we have

$Q = 40(950)(680 - 32) + 40(450 \times 10^3)$

$Q = 24624 kJ + 18000 kJ$

$Q = 42624 kJ$

Since this is the amount of aluminium per hour

so power required to melt is given by

$P = \frac{Q}{t}$

$P = \frac{42624}{3600} kW$

$P = 11.84 kW$

Since the efficiency is 85% so actual power required will be

$P = \frac{11.84}{0.85} = 13.93 kW$

Part B)

Total energy consumed by the furnace for 30 hours

$Energy = power \times time$

$Energy = 13.93 kW\times 30 h$

$Energy = 417.9 kWh$

now the total cost of energy consumption is given as

$R = P \times 20 \frac{Cents}{kWh}$

$R = 417.9 kWh\times 20 \frac{cents}{kWh}$

$R = 8357.6 Cents$

3 0

3 years ago