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Ivenika [448]
3 years ago
7

A large piece of jewelry has a mass of 130.8 g. A graduated cylinder initially contains 47.7 mL water. When the jewelry is subme

rged in the graduated cylinder, the total volume increases to 62.4 mL. (a) Determine the density (in g/cm3) of this piece of jewelry. g/cm3
Physics
1 answer:
Shkiper50 [21]3 years ago
8 0

Answer: The density of this piece of jewelry is 8.90g/cm^3

Explanation:

To calculate the density, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Mass of piece of jewellery = 130.8 g

Density of piece of jewellery = ?

Volume of piece of jewellery =( 62.4-47.7 ) ml = 14.7 ml = 14.7cm^3   1cm^3=1ml

Putting values in above equation, we get:

\text{Density of piece of jewellery}=\frac{130.8g}{14.7cm^3}=8.90g/cm^3

Thus density of this piece of jewelry is 8.90g/cm^3

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A bicyclist bikes the 90 mi to a city averaging a certain speed. The return trip is made at a speed that is 1 mph slower. Total
Lynna [10]

Answer:

his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph

Explanation:

Let say the speed of the bicycle while he moves towards the city is "v"

now the speed of the round trip must be smaller by 1 mph

so its speed for round trip will be

v_2 = v - 1

now we know that total time of the motion is 19 hr

so we will have

t_1 = \frac{90}{v}

t_2 = \frac{90}{v - 1}

so we will have

t_1 + t_2 = 19 hr

\frac{90}{v} + \frac{90}{v-1} = 19

90(2v - 1) = 19(v^2 - v)

19 v^2 - 199 v + 90 = 0

by solving above equation we have

v = 10 mph

so his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph

5 0
3 years ago
When a cup is placed on a table, which force prevents the cup from falling to the ground?
zhuklara [117]

Answer:

B. normal force

Explanation:

Because there is no frictional or resistance force. However gravitational force is applied downroad from the center of the cup thus the contact force that is perpendicular to the surface that an object contacts which is the normal force exerted upward from the table that prevents an object from falling.

6 0
4 years ago
a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the
tatuchka [14]
The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N
the force of the wind F, acting horizontally, with intensity
F=12 N
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
T \cos \alpha -F=0
T \sin \alpha -W=
By dividing the second equation by the first one, we get
\tan \alpha =  \frac{W}{F}= \frac{24.5 N}{12 N}=2.04
From which we find
\alpha = 63.8 ^{\circ}
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N




7 0
3 years ago
The angular velocity (in rpm) of the blade of a blender is given in. If θ=0rad at t=0s, what is the blade's angular position at
Liula [17]

Answer:

θ = 20.9 rad

Explanation:

In a blender after a short period of acceleration the blade is kept at a constant angular velocity, for which we can use the relationship

           w = θ / t

           θ = w t

if we know the value of the angular velocity we can find the angular position, we must remember that all the angles must be in radians

suppose that the angular velocity is w = 10 rpm, let us reduce to the SI system

          w = 10 rpm (\frac{2\pi  rad}{1 rev}) ( \frac{1 min}{60 s} )

= 1,047 rads

let's calculate

          θ = 1,047 20

          θ = 20.9 rad

7 0
3 years ago
Sally takes two bar magnets and randomly tapes one end of each bar magnet. she labels the magnets A and B. She brings the taped
Artyom0805 [142]
I already answered this quesiton. The fact is that there are only two kind of poles and since the two taped poles of the magnets labeled A and B attracts one to each other, we know that the two taped poles of the first two magnets are oppsosite.

Then, the taped pole of the third magnet has to be equal to one of the first two taped poles and opposite to the other of the first two taped poles.

That drives you to conclude (predict)  that when she brings the taped end of the third magnet (magnet C) near each of the first two magntes, in one case they will attract each other and in the other case they will repele mutually.

4 0
3 years ago
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