All categories

3 years ago

A block of metal is 3 cm on each side. It has a mass of 283.5 grams. What is the density of the metal.

1 answer:

5 0

$Density = \frac{Mass}{Volume}
\\\rho=\frac{m}{V}
\\\\\rho=\frac{283.5}{3^3}
\\\\\rho=\frac{283.5}{27}
\\\\\rho = 10.5 g/cm^3
$

You might be interested in

An object with a mass of 2.0 kg is accelerated at 5.0 m/s/s. the net force acting on the mass is

Vitek1552 [10]

First we need to know the equation:
F= Mass times acceleration
F = 2.0 kg times 5.0 m/s^2
multiply them to get the net force!
F= 10 N
N is newton
Hope this heps

7 0

3 years ago

Read 2 more answers

If the voltage in a circuit is 24 V and the current is 2 A, what is the total power in the circuit?

Drupady [299]

Maybe you can divide the volts its twelve if you do that but itll show you how much to double it by

5 0

3 years ago

Read 2 more answers

A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.

Mama L [17]

Answer: $f_{r}$ = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

$W_{x}$ = horizontal component of the weight = mgsinФ

$W_{y}$ = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - $f_{r}$ = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8 $m/s^{2}$

$f_{r}$ = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

$v^{2} = u^{2} + 2aS$

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

$2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}$

we slot in a into the equation below to get frictional force

mgsinФ - $f_{r}$ = ma

3 * 9.8 * sin 37 - $f_{r}$ = 3* 0.4

17.9633 - $f_{r}$ = 1.2

$f_{r}$ = 17.9633 - 1.2

$f_{r}$ = 16.49N

4 0

3 years ago

A supersonic jet, with a mass of 21,000 kg, departs from its home airbase with a velocity of 400 m/s due east. What is the jet’s

disa [49]

Answer:

Momentum P is 840000kgm/s or 8.4 × 10^6

Explanation:

Data :

Mass = 21000 kg

Velocity = 400 m/s

So momentum is given as

P = mv

P = 21000×400

P = 8400000 kgm/s

P = 8.4 × 10^6

6 0

2 years ago

Calculate the kinetic energy in joules of a 1,500 kg car that is moving at a speed of 42 km/h

Rzqust [24]

Data:
KE (Kinetic Energy) = ? (Joule)
m (mass) = 1500 Kg
v (speed) = 42 Km/h
converting to m/s (42 / 3.6), we have: v (speed) = 11.6 m/s

Formula:
$K_{E} = \frac{1}{2} m*v^2$

Solving:
$K_{E} = \frac{1}{2} m*v^2$
$K_{E} = \frac{1}{2} *1500*(11.6)^2$
$K_{E} = \frac{1}{2} *1500*134.56$
$K_{E} = \frac{201840}{2}$
$\boxed{\boxed{K_{E} = 100920\:Joule}}\end{array}}\qquad\quad\checkmark$

4 0

3 years ago