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3 years ago

Evaluate (5.7x106 kg) * (6.3x10-2 m/s2) and express the answer in scientific notation.

Physics

1 answer:

VladimirAG [237]3 years ago

5 0

Answer:

Force = 3.591 * 10⁵ Newton

Explanation:

Given the following data;

Mass = 5.7 * 10⁶ kg

Acceleration = 6.3 * 10^-2 m/s²

To express the force as a scientific notation;

Force = mass * acceleration

Force = 5.7 * 10⁶ * 6.3 * 10^-2

Force = 359100 Newton

In scientific notation;

A number is said to be written in scientific notation when we multiply the numbers between 1 and 10 by a power of 10.

Force = 3.591 * 10⁵ Newton

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In-s [12.5K]

Explanation:

the answer is y=strong winds move sand dunes in the desert

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3 years ago

Read 2 more answers

Hi please answer and show your work

jek_recluse [69]

Answer:

$\huge\boxed{\sf P.E. = 240\ MJ}$

$\huge\boxed{\sf K.E. = 19.6\ MJ}$

Explanation:

<u>Given:</u>

Mass = m = 200,000 kg

Vertical Distance = h = 120 m

Speed = v = 14 m/s

Acceleration due to gravity = g = 10 m/s²

<u>Required:</u>

1) Gravitational Potential Energy = P.E = ?

2) Kinetic Energy = K.E. = ?

<u>Formula:</u>

1) P.E. = mgh

2) K.E. = $\displaystyle \frac{1}{2} mv^2$

<u>Solution:</u>

1) P.E. = (200,000)(10)(120)

P.E. = 240,000,000 Joules

P.E. = 240 Mega Joules

P.E. = 240 MJ

2) K.E. = 1/2 (200000)(14)^2

K.E. = (100000)(196)

K.E. = 19,600,000 Joules

K.E. = 19.6 MJ

$\rule[225]{225}{2}$

Hope this helped!

4 0

3 years ago

A 150 kg car hits a wall with 45 N of force, what was its acceleration?

alisha [4.7K]

Answer:

a = 0.3 m/s²

Explanation:

Given: 45 N, 150 kg

To find: a

Formula: $a = \frac{F}{m}$

Solution: To find a, divide the force by the weight

A = F ÷ m

= 45 ÷ 150

= 0.3 m/s²

Newtons are derived units, equal to 1 kg-m/s². In other words, a single Newton is equal to the force needed to accelerate one kilogram one meter per second squared.

4 0

3 years ago

A tank having a volume of 0.85 m 3 initially contains water as a two-phase liquid-vapor mixture at 260 o C and a quality of 0.7.

Keith_Richards [23]

Answer:=14,160 kJ

Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below

Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase

C MPa m^3/kg kJ/kg kJ/kg

1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture

2 260 4.689 0.0422 2599 2797 1 Saturated Vapor

The mass initially contained in the tank is m1 = V/v1

m1 =0.85 m^3 /0.02993 m^3 /kg

= 28.4 kg

The mass finally contained in the tank is

m2 =V2/v

= 0.85 m^3 /0.0422 m^3 /kg

= 20.14 kg

The heat transfer is then

Qcv = m2u2 − m1u1 − he(m2 − m1)

Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ

4 0

3 years ago

To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3

hodyreva [135]

Answer : The correct option is, (c) $3.7\times 10^2J/^oC$

Explanation :

First we have to calculate the energy or heat.

Formula used :

$E=V\times I\times t$

where,

E = energy (in joules)

V = voltage (in volt)

I = current (in ampere)

t = time (in seconds)

Now put all the given values in the above formula, we get:

$E=(3.6V)\times (2.6A)\times (350s)$

$E=3276J$

Now we have to calculate the heat capacity of the calorimeter.

Formula used :

$C=\frac{E}{\Delta T}=\frac{E}{T_{final}-T_{initial}}$

where,

C = heat capacity of the calorimeter

$T_{initial}$ = initial temperature = $20.3^oC$

$T_{final}$ = final temperature = $29.1^oC$

Now put all the given values in this formula, we get:

$C=\frac{3276J}{(29.1-20.3)^oC}$

$C=372.27J/^oC=3.7\times 10^2J/^oC$

Therefore, the heat capacity of the calorimeter is, $3.7\times 10^2J/^oC$

7 0

3 years ago