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ioda
2 years ago
11

A dart gun shoots a dart with an angle of 45' above horizontal During the upward part of the trajectory the gravitational accele

ration g is directed 1 Upward and ts value changes with height 2 Upward and its value changes with height 3 There is no accelaration 4 Downward and its value is constant 5 Downward and ts value changes with height
Physics
1 answer:
ludmilkaskok [199]2 years ago
8 0

Answer:

4. Downward and its value is constant

Explanation:

As this is a case of projectile motion, we use the reference frame where upward direction to be positive for y, and in the same way to be negative in the downward direction. On another hand, we have that gravity is always acting this means that gravitational acceleration g is directed downward constantly over the dart not only during the upward but also during the downward part of the trajectory. And it is ruled by the following equations.

For the x-axis

v_{x}=v_{0}cos(45\°)=constant

x=(v_{0}cos(45\°))t

For the y-axis

v_{y}=v_{0}sin(45\°)-gt

y=v_{0}sin(45\°)t-\frac{1}{2} gt^{2}

Where v_{0}, is the initial velocity.

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Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00
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Answer:

2.62898\times 10^{-6}\ C/m^3

1979.99974\ N/C

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Electric field is given by

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\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3

The volume charge density for the sphere is 2.62898\times 10^{-6}\ C/m^3

E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C

The magnitude of the electric field is 1979.99974\ N/C

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