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ki77a [65]
2 years ago
11

If a book has a a mass of 2 kg, how much does the book weigh?

Physics
2 answers:
larisa86 [58]2 years ago
7 0
B. 19 N I got it here
Inessa05 [86]2 years ago
6 0

Answer:

A. 20 N

Explanation:

Weight of the book = mass × acceleration due to gravity

= 2 × 10

= 20 N

You might be interested in
I. What is the initial velocity of the car?
qaws [65]

Answer:

I. 0 m/s

II. 20 m/s

III. Part BC

Explanation:

I. Determination of the initial velocity.

From the diagram given above,

The motion of the car starts from the origin. This implies that the car start from rest and as such, the initial velocity of the car is 0 m/s

II. Determination of the maximum velocity attained.

From the diagram given above, we can see clearly that the maximum velocity is 20 m/s.

III. Determination of the part of the graph that represents zero acceleration.

It important that we know the meaning of zero acceleration.

Zero acceleration simply means the car is not accelerating. This can only be true when the car is moving with a constant velocity.

From the graph given above, the car has a constant velocity between B and C.

Therefore, part BC illustrates zero acceleration.

6 0
3 years ago
Usain Bolt's world-record 100 m sprint on August 16, 2009, has been analyzed in detail. At the start of the race, the 94.0 kg Bo
ZanzabumX [31]

a) 893 N

b) 8.5 m/s

c) 3816 W

d) 69780 J

e) 8030 W

Explanation:

a)

The net force acting on Bolt during the acceleration phase can be written using Newton's second law of motion:

F_{net}=ma

where

m is Bolt's mass

a is the acceleration

In the first 0.890 s of motion, we have

m = 94.0 kg (Bolt's mass)

a=9.50 m/s^2 (acceleration)

So, the net force is

F_{net}=(94.0)(9.50)=893 N

And according to Newton's third law of motion, this force is equivalent to the force exerted by Bolt on the ground (because they form an action-reaction pair).

b)

Since Bolt's motion is a uniformly accelerated motion, we can find his final speed by using the following suvat equation:

v=u+at

where

v is the  final speed

u is the initial speed

a is the acceleration

t is the time

In the first phase of Bolt's race we have:

u = 0 m/s (he starts from rest)

a=9.50 m/s^2 (acceleration)

t = 0.890 s (duration of the first phase)

Solving for v,

v=0+(9.50)(0.890)=8.5 m/s

c)

First of all, we can calculate the work done by Bolt to accelerate to a speed of

v = 8.5 m/s

According to the work-energy theorem, the work done is equal to the change in kinetic energy, so

W=K_f - K_i = \frac{1}{2}mv^2-0

where

m = 94.0 kg is Bolt's mass

v = 8.5 m/s is Bolt's final speed after the first phase

K_i = 0 J is the initial kinetic energy

So the work done is

W=\frac{1}{2}(94.0)(8.5)^2=3396 J

The power expended is given by

P=\frac{W}{t}

where

t = 0.890 s is the time elapsed

Substituting,

P=\frac{3396}{0.890}=3816 W

d)

First of all, we need to find what is the average force exerted by Bolt during the remaining 8.69 s of motion.

In the first 0.890 s, the force exerted was

F_1=893 N

We know that the average force for the whole race is

F_{avg}=820 N

Which can be rewritten as

F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}

And solving for F_2, we find the average force exerted by Bolt on the ground during the second phase:

F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}\\F_2=\frac{(0.890+8.69)F_{avg}-0.890F_1}{8.69}=812.5 N

The net force exerted by Bolt during the second phase can be written as

F_{net}=F_2-D (1)

where D is the air drag.

The net force can also be rewritten as

F_{net}=ma

where

a=\frac{v-u}{t} is the acceleration in the second phase, with

u = 8.5 m/s is the initial speed

v = 12.4 m/s is the final speed

t = 8.69 t is the time elapsed

Substituting,

a=\frac{12.4-8.5}{8.69}=0.45 m/s^2

So we can now find the average drag force from (1):

D=F_2-F_{net}=F_2-ma=812.5 - (94.0)(0.45)=770.2 N

So the increase in Bolt's internal energy is just equal to the work done by the drag force, so:

\Delta E=W=Ds

where

d is Bolt's displacement in the second part, which can be found by using suvat equation:

s=\frac{v^2-u^2}{2a}=\frac{12.4^2-8.5^2}{2(0.45)}=90.6 m

And so,

\Delta E=Ds=(770.2)(90.6)=69780 J

e)

The power that Bolt must expend just to voercome the drag force is given by

P=\frac{\Delta E}{t}

where

\Delta E is the increase in internal energy due to the air drag

t is the time elapsed

Here we have:

\Delta E=69780 J

t = 8.69 s is the time elapsed

Substituting,

P=\frac{69780}{8.69}=8030 W

And we see that it is about twice larger than the power calculated in part c.

3 0
3 years ago
A string has an abrupt change in linear density at its midpoint so that the speed of a pulse on the left side is 2/3 of that on
goldenfox [79]

Answer:

Explanation:

a) Linear density is greater on the left side because the velocity of wave in a string is inversely proportional to the linear mass density of the string

b) We should start pulse from the left hand side so the reflected wave does not get inverted because the wave traveling from the denser to lighter medium gets reflected in the same phase.

3 0
3 years ago
Part A What will be the equilibrium temperature when a 227 g block of copper at 283 °C is placed in a 155 g aluminum calorimeter
stellarik [79]

Answer:

T = 20.84°C

Explanation:

From the law of conservation of energy:

Heat Lost by Copper Block = Heat Gained by Aluminum Calorimeter + Heat Gained by Water

m_cC_c\Delta T_c = m_wC_w\Delta T_w + m_aC_a\Delta T_a

where,

m_c = mass of copper = 227 g

m_w = mass of water = 844 g

m_a = mass of aluminum = 155 g

C_c = specific heat capacity of calorimeter = 385 J/kg.°C

C_w = specific heat capacity of water = 4200 J/kg.°C

C_a = specific heat capacity of aluminum = 890 J/kg.°C

\Delta T_c = change in temperature of copper = 283°C - T

\Delta T_w = change in temperature of water = T - 14.6°C

\Delta T_a = change in temperature of aluminum = T - 14.6°C

T = equilibrium temperature = ?

Therefore,

(227\ g)(385\ J/kg.^oC)(283^oC-T)=(844\ g)(4200\ J/kg.^oC)(T-14.6^oC)+(155\ g)(890\ J/kg.^oC)(T-14.6^oC)\\\\24732785\ J - (87395\ J/^oC) T = (3544800\ J/^oC) T - 51754080\ J+ (137950\ J/^oC) T-2014070\ J\\\\24732785\ J +51754080\ J+2014070\ J = (3544800\ J/^oC) T+(137950\ J/^oC+(87395\ J/^oC) T\\\\78560935\ J = (3770145\ J/^oC) T\\\\T = \frac{78560935\ J}{3770145\ J/^oC}

<u>T = 20.84°C</u>

8 0
2 years ago
What is the voltage in a circuit that has a current of 10.0 amps and a resistance of 28.5 ohms?
marin [14]
Voltage = Current (I) × Resistance (R)
V = 10 × 28.5 = 285v
6 0
3 years ago
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