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jenyasd209 [6]
3 years ago
13

A projectile fired into the air, explodes, and splits into two halves of equal mass that hit the ground at the same time. If the

projectile had not exploded, it would have landed at point X, which is a distance R to the right of the launch point. After the explosion, one of the halves lands at point Y, which is a distance 2R to the right of the launch point. If air resistance is negligible, where did the other half land
Physics
1 answer:
steposvetlana [31]3 years ago
4 0

Answer:

At the Launch point

Explanation:

We are told that after the explosion, one of the halves landed at the point Y, which is a distance of 2R to the right side of the launch point.

This means that the only way one half of the object would have gone double the original distance of R is if it's velocity was two times that of the initial velocity. Furthermore, the only way that will make sense is if the other half fell straight down. So In conclusion, if air resistance is neglected, the other half will land at the initial velocity point which is at launch point.

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Answer:

ΔT = 1.22*10^-3 °C

Explanation:

First, you calculate the potential energy of the bird when it is at 35 m high. The potential energy is also the mechanical energy of the bird in this case.

U=mgh

m: mass of the bird = 0.75kg

g: gravitational constant = 9.8m/s^2

h: height = 35m

U=(0.75kg)(9.8m/s^2)(35m)=257.25\ J

All this energy is given to the water. You use the following formula in order to calculate the change in temperature:

Q=mc\Delta T

m: mass of the water = 50kg

c: specific heat of water = 4186 J/kg°C

Q is equal to U (potential energy of the bird) because the bird gives all its energy to water. By doing ΔT the subject of the formula you obtain:

\Delta T=\frac{Q}{mc}=\frac{257.25J}{(50kg)(4186J/kg°C)}=1.22*10^{-3}\ \°C

hence, the maximum rise in temperature is 0.00122 °C

7 0
3 years ago
Ask Your Teacher In Example 24.6, we found that the electric field of a charged disk approaches that of a charged particle for d
Ulleksa [173]

Answer: E = 7394.6N/C

Explanation:

Please find the attached file for the solution

8 0
3 years ago
A forklift lifts 5 boxes from the ground to a height of 2 meters (m). The boxes push down with a force of 1000 newtons (N). How
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Hello!

Answer:

2000 J

Explanation

Work equation is expressed as:

W=F.d.Cos \alpha

Where:

F: Applied force
d: traveled distance
α: Angle between the direction of the force and the direction of the movement. (in this case, both of the direction are the same, so the angle is 0°)

By substituting:

F=1000N.2m.Cos(0)=2000N.m=2000 J

Have a nice day!
8 0
4 years ago
A 4kg watermelon is dropped from a height of 45m. What is the velocity of the watermelon just before it hits the ground?
Makovka662 [10]

Answer:

v=30 m/s

Explanation:

h - height

g - acceleration due to gravity=10

t - time

v- velocity

h =  \frac{1}{2}  \times g \times t {}^{2}

45 = 5t²

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v=g×t

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3 0
3 years ago
Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B​
WARRIOR [948]

\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}

Explanation:

Given:

\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}

\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}

The cross product \textbf{A}×\textbf{B} is given by

\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|

=  \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}

= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}

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3 years ago
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