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kotykmax [81]
3 years ago
13

An object is thrown straight down with an initial speed of 4 m/s from a window which is 8 m above the ground. Calculate: a) The

time it takes the object to reach the ground b) Its velocity when it reaches the ground.
Physics
1 answer:
frosja888 [35]3 years ago
8 0

Answer:

0.41s

8.01m/s

Explanation:

Using the formula; v = u + at

Where;

u = initial velocity (m/s)

v = final velocity (m/s)

t = time (s)

a = acceleration (m/s²)

According to the provided information, u = 4m/s, s = 8m,

V = u + at

4 = 0 + 9.8t

4 = 9.8t

t = 0.41s

b) v = u + at

v = 4 + 9.8(0.41)

v = 4 + 4.018

v = 8.018m/s

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A delivery truck travels 2.8 km North, 1.0 km East, and 1.6 km South. The final displacement from the origin is ___km to the ___
34kurt

Answer:

The final displacement from the origin is <u>1.6</u> km to the <u>NE</u>

Explanation:

The directions in which the delivery truck travels are;

1) 2.8 km North = 2.8·\hat j, in vector form

2) 1.0 km East = 1.0·\hat i, in vector form

3) 1.6 km South = -1.6·\hat j, in vector form

Therefore, to find the final displacement, Δx, of the delivery truck, we add the individual displacements as follows;

Final displacement, Δd = 2.8·\hat j + 1.0·\hat i +(-1.6·\hat j) = 1.2·\hat j + 1.0·\hat i

Final displacement, = 1.0·\hat i + 1.2·\hat j

Where;

Δx = The displacement in the x-direction = 1.0·\hat i

Δy = The displacement in the y-direction = 1.2·\hat j

The magnitude of the resultant displacement vector is given as follows

\left | d \right | = √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)

The magnitude of the resultant displacement vector ≈ 1.6 km

The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE

Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.

3 0
3 years ago
What is a gravitational wave and why was it so hard to detect?
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5 0
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You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene
uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is 150\,^{\circ}C. According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T) (1)

Where:

\rho_{k} - Density of kerosene, measured in kilograms per cubic meter.

V_{k} - Volume of kerosene, measured in cubic meters.

c_{k}, c_{t} - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

T_{k,o}, T_{t,o} - Initial temperatures of kerosene and tin, measured in degrees Celsius.

T - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}

If we know that m_{t} = 1.83\,kg, c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}, T_{t,o} = 88\,^{\circ}C, T_{k,o} = 24.0\,^{\circ}C, T = 57\,^{\circ}C, c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C} and \rho_{k} = 820\,\frac{kg}{m^{3}}, then the volume of the liquid needed to accomplish this task without boiling is:

V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}

V_{k} = 2.273\times 10^{-4}\,m^{3}

V_{k} = 0.273\,L

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A) 5.0\cdot 10^{-11} m

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The energy of a photon is related to its wavelength by the equation

E=\frac{hc}{\lambda}

where

h=6.63\cdot 10^{-34}Js is the Planck constant

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\lambda is the wavelength

Re-arranging the equation for the wavelength, we find

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4\cdot 10^{-15}J}=5.0\cdot 10^{-11} m

B) 2.0\cdot 10^{-11} m

The energy of an x-ray photon used in microtomography is 2.5 times greater than the energy of the photon used in part A), so its energy is

E=2.5 \cdot (4\cdot 10^{-15}J)=1\cdot 10^{-14} J

And so, by using the same formula we used in part A), we can calculate the corresponding wavelength:

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1\cdot 10^{-14}J}=2.0\cdot 10^{-11} m

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