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3 years ago

11

A tornado lifts a truck 252 m above the ground. As the storm continues, the tornado throws the truck horizontally. It lands 560

m away from where it was picked up.
How fast was the truck traveling horizontally through its flight? Round to the nearest tenth.

2 answers:

DochEvi [55]3 years ago

7 0

Answer;

-78.1 m/s

Explanation;

-Vertical component of velocity is parabolic while the Horizontal component is constant.

Initial velocity = 0 , initial height = 252

dy = v0*t + ½at²

dy= vo + 0.5gt^2

252 = 0 + 4.9t^2

t = 7.17 seconds

horizontal:

vx = (dx)/t = (560 m)(7.17 s) = 78.103 m/s

To the nearest tenth; 78.1 M/s

Rama09 [41]3 years ago

6 0

The answer is 78.1 I already checked it

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Cole is riding a sled with initial speed of 5 m/s from west to east. the frictional force of 50 n exists due west. the mass of t

stepan [7]

We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
$F=ma$
where $F=-50 N$ is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
$a= \frac{F}{m}= \frac{-50 N}{100 kg}=-0.5 m/s^2$

Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
$a= \frac{v_f^2-v_i^2}{2d}$
where
$v_f=0$ is the final speed of the sled
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By rearranging the equation, we find d:
$d= \frac{v_f^2-v_i^2}{2a}= \frac{-(5 m/s)^2}{2 \cdot (-0.5 m/s^2)}=25 m$

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3 years ago

A visible light has a wavelength of 727.3 nm. Determine its frequency, energy per photon, and color.

Snezhnost [94]

Answer:

$f=4.12\times 10^{14}\ Hz$ and $E=2.73\times 10^{-19}\ J$

Explanation:

The wavelength of a visible light is 727.3 nm.

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The formula is as follows :

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f is the frequency of the visible light

$f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{727.3 \times 10^{-9}}\\\\f=4.12\times 10^{14}\ Hz$

Energy of a photon is given by :

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Red color has a frequency of $4.12\times 10^{14}\ Hz$ and energy per photon is $2.73\times 10^{-19}\ J$ .

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schepotkina [342]

Answer:

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2 years ago

The force F required to compress a spring a distance x is given by F 2 F0 5 kx where k is the spring constant and F0 is the prel

IrinaVladis [17]

Answer:

a)W=8.333lbf.ft

b)W=0.0107 Btu.

Explanation:

<u>Complete question</u>

The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.

Solution

Preload = F₀=0 lbf

Spring constant k= 200 lbf/in

Initial length of spring x₁=0

Final length of spring x₂= 1 in

At any point, the force during deflection of a spring is given by;

F= F₀× kx where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.

$W=\int\limits^2_1 {} \, Fds \\\\\\W=\int\limits^2_1( {F_0+kx} \,) dx \\\\\\W=\int\limits^a_b {kx} \, dx ; F_0=0\\\\\\W=k\int\limits^2_1 {x} \, dx \\\\\\W=k*\frac{1}{2} (x_2^{2}-x_1^{2} )\\\\\\W=200*\frac{1}{2} (1^2-0)\\\\\\W=100.lbf.in\\\\$

Change to lbf.ft by dividing the value by 12 because 1ft=12 in

100/12 = 8.333 lbf.ft

work required to compress the spring, W=8.333lbf.ft

The work required to compress the spring in Btu will be;

1 Btu= 778 lbf.ft

?= 8.333 lbf.ft----------------cross multiply

(8.333*1)/ 778 =0.0107 Btu.

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3 years ago

A ball with a mass of 2000 g is floating on the surface of a pool of water. What is the minimum volume that the ball could have

Doss [256]

Answer:

$2000\; {\rm cm^{3}}$ .

Explanation:

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The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.

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Let $g$ denote the gravitational field strength. The weight of this ball would be $m(\text{ball}) \, g$ . Likewise, the weight of water displaced would be $m(\text{water})\, g$ .

For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:

$\text{buoyancy} \ge m(\text{ball})\, g$ .

At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:

$\text{buoyancy} = m(\text{water}) \, g$ .

Therefore:

$m(\text{water})\, g = \text{buoyancy} \ge m(\text{ball})\, g$ .

$m(\text{water}) \ge m(\text{ball})$ .

In other words, the mass of water that this ball displaced should be greater than or equal to the mass of of the ball. Let $\rho(\text{water})$ denote the density of water. The volume of water that this ball should displace would be:

$\begin{aligned}V(\text{water}) &= \frac{m(\text{water})}{\rho(\text{water})} \\ &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \end{aligned}$ .

Given that $m(\text{ball}) = 2000\; {\rm g}$ while $\rho = 1.00\; {\rm g\cdot cm^{-3}}$ :

$\begin{aligned}V(\text{water}) &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \\ &= \frac{2000\; {\rm g}}{1.00\; {\rm g\cdot cm^{-3}}} \\ &= 2000\; {\rm cm^{3}}\end{aligned}$ .

In other words, for this ball to stay afloat, at least $2000\; {\rm cm^{3}}$ of the volume of this ball should be under water. Therefore, the volume of this ball should be at least $2000\; {\rm cm^{3}}\!$ .

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