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inessss [21]
3 years ago
15

Drugs do all of their work at the brain's neurons. or the spaces between

Physics
1 answer:
otez555 [7]3 years ago
4 0

Answer:

spaces between

Explanation:

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A basketball player throws a chall -1 kg up with an initial speed of his hand at shoulder height = 2.15 m Le gravitational poten
Talja [164]

Complete Question:

A basketball player tosses a basketball m=1kg straight up with an initial speed of v=7.5 m/s. He releases the ball at shoulder height h= 2.15m. Let gravitational potential energy be zero at ground level

a)  Give the total mechanical energy of the ball E in terms of maximum height hn it reaches, the mass m, and the gravitational acceleration g.

b) What is the height, hn in meters?

Answer:

a) Energy = mghₙ

b) Height, hₙ = 5.02 m

Explanation:

a) Total energy in terms of maximum height

Let maximum height be hₙ

At maximum height, velocity, V=0

Total mechanical energy , E = mgh + 1/2 mV^2

Since V=0 at maximum height, the total energy in terms of maximum height becomes

Energy = mghₙ

b) Height,  hₙ in meters

mghₙ = mgh + 1/2 mV^2

mghₙ = m(gh + 1/2 V^2)

Divide both sides by mg

hₙ = h + 0.5 (V^2)/g

h = 2.15m

g = 9.8 m/s^2

V = 7.5 m/s

hₙ = 2.15 + 0.5(7.5^2)/9.8

hₙ = 2.15 + 2.87

hₙ = 5.02 m

6 0
3 years ago
11. A candle is placed in front of a plane mirror. The calculated value of m,
PtichkaEL [24]

Answer: If the magnification is positive, the image is upright compared to the object; if m is negative, the image is inverted compared to the object.

(I think, I was also stuck on this question for a bit)

7 0
2 years ago
An object thrown in the air has a velocity after t seconds that can be described by v(t) = -9.8t + 24 (in meters/second) and a h
marin [14]

Answer and Explanation: Kinetic energy is related to movement: it is the energy an object possesses during the movement. it is calculated as:

K=\frac{1}{2}mv^{2}

For the object thrown in the air:

K=\frac{1}{2}.2.[v(t)]^{2}

K=(-9.8t+24)^{2}

K=96.04t^{2}-470.4t+576

Kinetic energy of the object as a function of time: K=96.04t^{2}-470.4t+576

Potential energy is the energy an object possesses due to its position in relation to other objects. It is calculated as:

U=mgh

For the object thrown in the air:

U=9.8.2.h(t)

U=9.8.2.(-4.9t^{2}+24t+60)

U=-96.04t^{2}+470.4t+1176

Potential energy as function of time: U=-96.04t^{2}+470.4t+1176

Total kinetic and potential energy, also known as mechanical energy is

TME = 96.04t^{2}-470.4t+576 + (-96.04t^{2}+470.4t+1176)

TME = 1752

The expression shows that total energy of an object thrown in the air is constant and independent of time.

6 0
2 years ago
How does the position of an object relate to the energy stored in an object?
Kamila [148]

Answer:

Potential Energy

Explanation:

Potential energy is the energy stored in an object due to it's position relative to some zero position. An object possesses gravitational potential energy if it is positioned at a height above (or below) the zero height.

4 0
2 years ago
Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha
Inessa05 [86]

Answer:

+1.11\mu C

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

Q_1 is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to Q_2 and Q_3 at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of Q_1.

Let the electric field intensity due to Q_2 be +E_2 and that due to Q_3 be -E_3 since the charge is negative. Hence at the origin;

+E_2-E_3=0..................(1)

From equation (1) above, we obtain the following;

E_2=E_3.................(2)

From Coulomb's law the following relationship holds;

+E_2=\frac{kQ_2}{r_2^2}\\  

-E_3=\frac{kQ_3}{r_3^2}

where r_2 is the distance of Q_2 from the origin, r_3 is the distance of Q_3 from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)

k can cancel out from both side of equation (3), so that we finally obtain the following;

\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)

Given;

Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m

Substituting these values into equation (4); we obtain the following;

\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\

Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C

6 0
3 years ago
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