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3 years ago

What’s the answer to this question

Physics

1 answer:

alexgriva [62]3 years ago

7 0

Answer:

space = 66.24 [m]

Explanation:

To solve this problem we must remember that the average speed is defined as the relationship between a space traveled over a certain time.

$Av = \frac{space}{time}$

where:

space [m]

Av = average velocity = 3.6 [m/s]

time = 18.4 [s]

$space = 3.6*18.4\\space = 66.24 [m]$

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If a cup of coffee is at 90°C and a person with a body temperature of 36°C touches it, how will heat flow between them?

Savatey [412]

Answer:

According to II law of thermodynamics the heat always transfers from high temperature to low temperature body without aid of any external energy. So, according to this law the heat transfers from cup of coffee to the person body until both bodies reaches to the equilibrium state.

After attaining the equilibrium state, the heat transfer may be ceases between the bodies.

Whenever the person touches the hot coffee, he senses the heat that means the heat is transferring in to his body.

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4 years ago

Read 2 more answers

A television camera is positioned 4000 ft from the base of a rocket launching pad. The angle of elevation of the camera has to c

Makovka662 [10]

Answer:

a ) 540 ft /s

b ) .144° /s

Explanation:

Let at any moment h be the height of the rocket . The distance of rocket from camera will be R

R ² = 4000² + h²

Differentiating both sides with respect to t

R dR/dt = h dh/dt

dR/dt = h/R dh/dt

a ) Given speed of rocket

dh/dt = 900

h = 3000

R² = 4000² + 3000²

R = 5000

dR/dt = h/R dh/dt

= (3000 / 5000 ) X 900

dR/dt = 540 ft /s

b ) Let θ be angle of elevation at the moment .

4000 / R = cosθ

Differentiating with respect to t

- 4000 x 1 / R² dR/dt = - sinθ dθ / dt

4000 x ( 1/ 5000² ) x 540 = 3 /5 x dθ / dt

.0864 = 3/5 dθ / dt

dθ / dt = .144° /s

8 0

3 years ago

A ball with mass m kg is thrown upward with initial velocity 22 m/s from the roof of a building 25 m high. Neglect air resistanc

makvit [3.9K]

Answer:

(a) 24.7 metres

(b) 5.4 seconds

Explanation:

(a) To solve this part, we only need the initial velocity of the ball, u = 22 m/s

We apply one of Newton's equations of motion to solve this:

v² = u² - 2gs

Where v = final velocity

u = initial velocity

g = acceleration due to gravity = 9.8 m/s

s = distance moved by ball

Note: The equation has a negative sign because the ball is moving against the gravitational force.

The maximum height reached by the ball will be attained when the ball's final velocity, v, is 0 m/s.

Hence:

0² = 22² - 2 * 9.8 * s

=> 19.6s = 22²

19.6s = 484

s = 484 / 19.6 = 24.7 m

The maximum height reached by the ball is 24.7 m

(b) We have to solve two parts, the time it take a takes the ball to reach its maximum height and the time it takes to fall after reaching maximum height.

TIME IT TAKES TO REACH MAXIMUM HEIGHT

Using another one of Newton's equations, we have that:

v = u - gt

We already have that v = 0 m/s and u = 22 m/s.

Hence:

0 = 22 - 9.8t

9.8t = 22

=> t = 22 /9.8 = 2.2 seconds

TIME IT TAKES TO FALL AFTER REACHING MAXIMUM HEIGHT

Since the ball was thrown from a height of 25 m and it has now reached 24.7 m higher, the ball is now at a height of:

25 + 24.7 = 49.7 m

We apply another one of Newton's equations of motion to obtain the time it will take to reach the ground:

s = ut + ½gt²

t = time taken

The initial velocity will become zero (0 m/s) at the point because the ball is beginning its descent.

Hence:

49.7 = (0*t) + ½ * 9.8 * t²

49.7 = 4.9t²

=> t² = 49.7 / 4.9 = 10.14

=> t = 3.2 seconds

It will take the ball 3.2 seconds to hit the ground after it starts descending.

The total time it takes the ball to hit the ground will therefore be:

T = 2.2 + 3.2 = 5.4 seconds

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3 years ago

Read 2 more answers

Determine whether the following elements are more likely to be a metal or nonmetal

olga2289 [7]

A) Metal
B) Non-metal
C) Non-metal
D) Metal

6 0

3 years ago

After fracture, the total length was 47.42 mm and the diameter was 18.35 mm. Plot the engineering stress strain curve and calcul

Fittoniya [83]

Answer:

Part a: <em>The value of yield strength at 0.2% offset is obtained from the engineering stress-strain curve which is given as 274 MPa.</em>

Part b: <em>The value of tensile strength is obtained from the engineering stress-strain curve which is given as 417 MPa</em>

Part c: <em>The value of Young's modulus at given point is 172 GPa.</em>

Part d: <em>The percentage elongation is 18.55%.</em>

Part e: The percentage reduction in area is 15.81%

Explanation:

From the complete question the data is provided for various Loads in ductile testing machine for a sample of d0=20 mm and l0=40mm. The plot is drawn between stress and strain whose values are calculated using following formulae. The corresponding values are attached with the solution.

The engineering-stress is given as

$\sigma=\frac{F}{A}\\\sigma=\frac{F}{\pi \frac{d_0^2}{4}}\\\sigma=\frac{F}{\pi \frac{(20 \times 10^-3)^2}{4}}\\\sigma=\frac{F}{3.14 \times 10^{-4}}$

Here F are different values of the load

Now Strain is given as

$\epsilon=\frac{l-l_0}{l_0}\\\epsilon=\frac{\Delta l}{40}\\$

So the curve is plotted and is attached.

<em>The value of yield strength at 0.2% offset is obtained from the engineering stress-strain curve which is given as 274 MPa.</em>

<em>The value of tensile strength is obtained from the engineering stress-strain curve which is given as 417 MPa</em>

Young's Modulus is given as

$E=\frac{\sigma}{\epsilon}\\E=\frac{238 /times 10^6}{0.00138}\\E=172,000 MPa\\E=172 GPa$

<em>The value of Young's modulus at given point is 172 GPa.</em>

The percentage elongation is given as

$Elongation=\frac{l_f-l_0}{l_0} \times 100\\Elongation=\frac{47.42-40}{40}\times 100\\Elongation=18.55 \%\\$

So the percentage elongation is 18.55%

The reduction in area is given as

$Reduction=\frac{A_0-A_n}{A_0} \times 100\\Reduction=\frac{\pi \frac{d_0^2}{4}-\pi \frac{d_n^2}{4}}{\pi \frac{d_0^2}{4}}\times 100\\Reduction=\frac{{d_0^2}-{d_n^2}}{{d_0^2}} \times 100\\Reduction=\frac{{20^2}-{18.35^2}}{{20^2}} \times 100\\Reduction=15.81\%$

So the reduction in area is 15.81%

6 0

3 years ago