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docker41 [41]
3 years ago
12

two objects are thrown from the top of a tall building and experience no appreciable air resistance. one is thrown up and the ot

her is thrown down, both with the same initial speed. what are their speeds when they hit the street?
Physics
2 answers:
NNADVOKAT [17]3 years ago
5 0

Answer:

Final speeds are equal to each other.

Explanation:

Both objects experiment a change in their velocities because of gravity. Let model each object by following equations:

Object A - Throw Up

v = -\sqrt{v_{o}^{2}+2\cdot g \cdot h}

Object B - Throw Down

v = -\sqrt{v_{o}^{2}+2\cdot g \cdot h}

Where h is the height difference between the top of the building and the street. Let consider upward speed positive.

Final speeds are equal to each other.

Iteru [2.4K]3 years ago
3 0

Answer:

The two objects are traveling at the same speed.

Explanation:

Neglecting air resistance, an object that is thrown up from the top of a tall building has the same speed as the second object thrown down from the top of the same tall building since the initial speed is the same.

The object thrown up is not traveling faster neither is the object thrown down traveling faster.  

Therefore, the two objects will have the same speed when they hit the ground but their time of landing might be different.

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Most automobiles have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made o
Colt1911 [192]

Answer:

There is a loss of fluid in the  container of 0.475L

Explanation:

To solve the problem it is necessary to take into account the concepts related to the change of voumen in a substance depending on the temperature.

The formula that describes this thermal expansion process is given by:

\Delta V = \beta V_0 \Delta T

Where,

\Delta V =Change in volume

V_0 =Initial Volume

\Delta T = Change in temperature

\beta = coefficient of volume expansion (Coefficient of copper and of the liquid for this case)

There are two types of materials in the container, liquid and copper, so we have to change the amount of Total Volume that would be subject to,

\Delta V_T = \Delta V_l - \Delta V_c

Where,

\Delta V_l= Change in the volume of liquid

\Delta V_c= Change in the volume of copper

Then replacing with the previous equation we have:

\Delta V = \beta_l V_0 \Delta T- \beta_c V_0 \Delta T

\Delta V = (\beta_l-\beta_c)V_0\Delta T

Our values are given as,

Thermal expansion coefficient for copper and the liquid to 20°C is

\beta_c = 51*10^{-6}/\°C

\beta_l = 400*10^{-6}/\°C

V_0 = 16L

\Delta T = (95\°C-10\°C)

Replacing we have that,

\Delta V = (\beta_l-\beta_c)V_0\Delta T

\Delta V = (400*10^{-6}/\°C-51*10^{-6}/\°C)(16L)(95\°C-10\°C)

\Delta V = 0.475L

Therefore there is a loss of fluid in the container of 0.475L

6 0
3 years ago
A mass attached to a 50.0 cm long string starts from rest and is rotated 40 times in one minute before reaching a final angular
ch4aika [34]

To solve this problem it is only necessary to apply the kinematic equations of angular motion description, for this purpose we know by definition that,

\theta = \frac{1}{2}\alpha t^2 +\omega_0 t + \theta_0

Where,

\theta = Angular Displacement

\alpha =Angular Acceleration

\omega_0 = Angular velocity

\theta_0 =Initial angular displacement

For this case we have neither angular velocity nor initial angular displacement, then

\theta = \frac{1}{2}\alpha t^2

Re-arrange for \alpha,

\alpha = \frac{2\theta}{t^2}

Replacing our values,

\alpha = \frac{2(40rev*\frac{2\pi rad}{1rev})}{60^2}

\alpha = 0.139rad/s

Therefore the ANgular acceleration of the mass is 0.139rad/s^2

4 0
3 years ago
A field measuring 12 meters by 16 meters is to have a brick paver walkway installed all around it, increasing the total area to
kolezko [41]

Answer:

1.5 m

Explanation:

Length. L = 12 m

Width, W = 16 m

Area, A = 12 x 16 = 192 m^2

Let the width of pavement be d.

The new length, L' = 12 + 2d

the new width, W' = 16 + 2d  

New Area, A' = L' x W' = (12 + 2d)(16 + 2d) = 192 + 56 d + 4d^2

Difference in area = A' - A

285 =  192 + 56 d + 4d^2 - 192

93 =  56 d + 4d^2

4d^2 + 56 d - 93 = 0

d = \frac{-56\pm \sqrt{56^{2}+4\times 4\times 93}}{8}

d=\frac{-56\pm 87.72}{8}\

d = 1.5 m

Thus, the width of the pavement is 1.5 m.

6 0
3 years ago
I need to show my work as well but on the computer so, please show work for how you got the answers. Thank you!
balandron [24]

Answer:

1) 1H_2 + 1Cl_2 => 2HCl

2) 2Al + 6HCl => 2AlCl_3 + 3H_2

3) 1Ca_2Br_2 + 2NaCO_3 => 2CaCO_3 + 2NaBr

4) 3NaOH + 1FeCl_3 => 3NaCl + 1Fe(OH)_3

Explanation:

4 0
3 years ago
PLEASEEE HELPPPP
solniwko [45]

Answer:

Explanation:

average speed more than 25.0m/s.

5 0
3 years ago
Read 2 more answers
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