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2 years ago

Determine the mass of the object below to the correct degree of precision.

Physics

1 answer:

dedylja [7]2 years ago

3 0

For this question place the object on the scale and move the slides to change the mass until it is level. That will tell you the mass.

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A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Complete the following statemen

Eva8 [605]

Answer:

Keeping the speed fixed and decreasing the radius by a factor of 4

Explanation:

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. The centripetal acceleration is given by :

$a=\dfrac{v^2}{R}$

We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"

It can be done by keeping the speed fixed and decreasing the radius by a factor of 4 such that,

R' = R/4

New centripetal acceleration will be,

$a'=\dfrac{v^2}{R'}$

$a'=\dfrac{v^2}{R/4}$

$a'=4\times \dfrac{v^2}{R}$

$a'=4\times a$

So, the centripetal acceleration of the ball can be increased by a factor of 4.

7 0

2 years ago

50 grams of ice cubes at -15°C are used to chill a water at 30°C with mass mH20 = 200 g. Assume that the water is kept in a foam

Arada [10]

Answer : The final temperature is, $25.0^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ice = $2.09J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ice = 50 g

$m_2$ = mass of water = 200 g

$T_f$ = final temperature = ?

$T_1$ = initial temperature of ice = $-15^oC$

$T_2$ = initial temperature of water = $30^oC$

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

$50g\times 2.09J/g^oC\times (T_f-(-15))^oC=-200g\times 4.184J/g^oC\times (T_f-30)^oC$

$T_f=25.0^oC$

Therefore, the final temperature is, $25.0^oC$

5 0

3 years ago

Starting from one oasis, a camel walks 25 km in a direction 30° south of west and then walks 30 km toward the north to a second

shepuryov [24]

Answer:

distance between both oasis ( 1 and 2) is 27.83 Km

Explanation:

let d is the distance between oasis1 and oasis 2

from figure

OC = 25cos 30

OE = 25sin30

OE = CD

Therefore BC = 30-25sin30

distance between both oasis ( 1 and 2) is calculated by using phytogoras theorem

in $\Delta BCO$

$OB^2 = BC^2 + OC^2$

PUTTING ALL VALUE IN ABOVE EQUATION

$d^2 = 930-25sin30)^2 + (25cos30)^2$

$d^2 = 775$

d = 27.83 Km

distance between both oasis ( 1 and 2) is 27.83 Km

3 0

2 years ago

Help me with this question Please

zysi [14]

Answer:

W has the lowest density and Y has the greatest density

Explanation:

Density of W = mass/volume = 11/24 = 0.45

Density of X = mass/volume = 11/12 = 0.91

Density of Y = m/v = 5.5/4 = 1.375

Density of Z = m/v = 5.5/11 = 0.5

From these we can find the answer......

Hope this answer is useful......

3 0

2 years ago

A plane drops a hamper of medical supplies from a height of 5000 m during a practice run over the ocean. The plane’s horizontal

Marizza181 [45]

Answer:

346.70015 m/s

Explanation:

In the x axis speed is

$v_x=149\ m/s$

In the y axis

$v_y=\sqrt{2gh}\\\Rightarrow v_y=\sqrt{2\times 9.8\times 5000}$

The resultant velocity is given by

$v=\sqrt{v_x^2+v_y^2}\\\Rightarrow v=\sqrt{149^2+2\times 9.8\times 5000}\\\Rightarrow v=346.70015\ m/s$

The magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean is 346.70015 m/s

4 0

2 years ago