The specific heat of water is 4,186 J/kg.'C. Approximately how much heat must

You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads 28.4 NN . If you t

shtirl [24]

Answer:

2491.23 kg/m³

Explanation:

From Archimedes principle,

R.d = weight of object in air/ upthrust in water = density of the object/density of water

⇒ W/U = D/D' ....................... Equation 1

Where W = weight of the ceramic statue, U = upthrust of the ceramic statue in water, D = density of the ceramic statue, D' = density of water.

Making D the subject of the equation,

D = D'(W/U).................... Equation 2

Given: W = 28.4 N, U = lost in weight = weight in air- weight in water

U = 28.4 - 17.0 = 11.4 N,

Constant: D' = 1000 kg/m³.

Substitute into equation 2,

D = 100(28.4/11.4)

D = 2491.23 kg/m³

Hence the density of the ceramic statue = 2491.23 kg/m³

7 0

3 years ago

A cyclist accelerates from a velocity of 10 miles/hour east until reaching a velocity of 20 miles/hour east in 5 seconds. What w

Sliva [168]

Answer:

$a = 0.894\ m/s^2$

Explanation:

<u>Motion with Constant Acceleration</u>

A body moves with constant acceleration when the speed changes uniformly in time. The equation used to find the final speed vf is

$v_f=v_o+at$

Where vo is the initial speed, a is the acceleration, and t is the time.

The cyclist has an initial speed of vo=10 miles/hour and ends up at vf=20 miles/hour in t=5 seconds.

Both speeds are given in miles/hour and we must convert it to m/s:

1 mile/hour = 0.44704 m/s

10 mile/hour = 4.47 m/s

20 mile/hour = 8.94 m/s

The acceleration is calculated by solving for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

$\displaystyle a=\frac{8.94-4.47}{5}$

$a = 0.894\ m/s^2$

3 0

2 years ago

A 40 g ball rolls around a 30 cm -diameter L-shaped track, shown in the figure, (Figure 1)at 60 rpm . What is the magnitude of t

levacccp [35]

Answer:

0.47 N

Explanation:

Here we have a ball in motion along a circular track.

For an object in circular motion, there is a force that "pulls" the object towards the centre of the circle, and this force is responsible for keeping the object in circular motion.

This force is called centripetal force, and its magnitude is given by:

$F=m\omega^2 r$

where

m is the mass of the object

$\omega$ is the angular velocity

r is the radius of the circle

For the ball in this problem we have:

m = 40 g = 0.04 kg is the mass of the ball

$\omega =60 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=6.28 rad/s$ is the angular velocity

r = 30 cm = 0.30 m is the radius of the circle

Substituting, we find the force:

$F=(0.040)(6.28)^2(0.30)=0.47 N$

3 0

2 years ago

On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3, length 88.8 cm and diameter 2.30 cm fro

vichka [17]

Answer:

w = 28.25 N

Explanation:

To do this, we need to use two expressions.

First, to calculate the weight of any object, we use the 2nd law of newton. In this case, the weight is:

w = m*g (1)

However we do not have the mass of the rod. We need to calculate that. To calculate the mass, we'll use the expression of density which is:

d = m/V

From here, we solve for mass:

m = d * V (2)

Finally, we can know the volume of the rod, because is cylindrical, therefore, the volume of a cylinder is:

V = π * r² * h (3)

So, in resume, we need to solve for the volume of the rod, then, the mass ans finally the weight. Let's calculate the volume of the rod, converting the units of centimeter to meters, just dividing by 100:

diameter = 2.3 cm ---> radius = 2.3/2 = 1.15 cm -----> 0.0115 m

Length or height = 88.8 cm ----> 0.888 m

Replacing in (3):

V = π * (0.0115)² * 0.888

V = 3.69x10⁻⁴ m

Now, let's use (2) to calculate the mass:

m = 7800 * 3.69x10⁻⁴

m = 2.88 kg

Finally for the weight, we'll use expression (1):

w = 2.88 * 9.81

<h2>w = 28.25 N</h2><h2>And this is the weight of the rod.</h2>

4 0

3 years ago

What kind of frequency do radio waves have?<br><br>A. High frequency <br><br>B. Low frequency

inn [45]

B low frequency it is the lowest frequency

7 0

3 years ago