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pashok25 [27]
4 years ago
15

.... Answer this... ​

Physics
1 answer:
AysviL [449]4 years ago
8 0

the ansnwer is ATP and ADP whitch is why we need food

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An electron moving along the x axis has an initial speed of 1 × 106 m/s at the origin. Its speed is reduced to 5 × 105 m/s at th
mina [271]

Answer:

-2.13 V

Explanation:

Given parameters are:

v_i = 1 * 10^6 m/s

v_f = 5 * 10^5 m/s

q = -e = 1.602 * 10^{-19} C

m = 9.109 * 10^{-31} kg

By conservation of energy principle, we know that the potential difference between two points is equal to the change in the kinetic energy between these points.

-\Delta U = \Delta KE\\\\-e\Delta V = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2

Hence,

1.602*10^{-19}\Delta V = \frac{1}{2}*9.11*10^{-31}((5*10^5)^2 - (1*10^6)^2)

⇒ \Delta V = -2.13 V

5 0
3 years ago
I have three questions. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s.
AfilCa [17]

1. 5.5 m/s

We can solve the problem by applying the law of conservation of momentum. The total momentum before the collision must be equal to the total momentum after the collision, so we have:

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where

m1 = 0.4 kg is the mass of the ball

u1 = 18 m/s is the initial velocity of the ball

m2 = 0.2 kg is the mass of the bottle

u2 = 0 is the initial velocity of the bottle (which is initially at rest)

v1 = ? is the final velocity of the ball

v2 = 25 m/s is the final velocity of the bottle

Substituting and re-arranging the equation, we can find the final velocity of the ball:

v_1 = \frac{m_1 u_1 - m_2 v_2}{m_1}=\frac{(0.4 kg)(18m/s)-(0.2 kg)(25 m/s)}{0.4 kg}=5.5 m/s


2. 22.2 m/s

We can solve the problem again by using the law of conservation of momentum; the only difference in this case is that the bullet and the block, after the collision, travel together at the same speed v. So we can write:

m_1 u_1 + m_2 u_2 = (m_1 +m_2) v

where

m1 = 0.04 kg is the mass of the bullet

u1 = 300 m/s is the initial velocity of the bullet

m2 = 0.5 kg is the mass of the block

u2 = 0 is the initial velocity of the block (which is initially at rest)

v = ? is the final velocity of the bullet+block, which stick and travel together

Substituting and re-arranging the equation, we can find the final velocity of bullet+block:

\frac{m_1 u_1}{m_1 +m_2}=\frac{(0.04 kg)(300 m/s)}{0.04 kg+0.5 kg}=22.2 m/s


3. 6560 N

The impulse exerted on the ball is equal to its change in momentum:

I=\Delta p (1)

The impulse can be rewritten as product between force and time of collision:

I=F \Delta t

while the change in momentum of the ball is equal to the product between its mass and the change in velocity:

\Delta p = m\Delta v = m(v_f -v_i)

So, eq.(1) becomes

F \Delta t = m(v_f -v_i)

where:

F = ? is the unknown force

\Delta t = 0.002 s is the duration of the impact

m = 0.16 kg is the mass of the ball

v_f = 44 m/s is the final velocity of the ball

v_i = -38 m/s is its initial velocity (we must add a negative sign, since it is in opposite direction to the final velocity)

So, by using the equation, we can find the force:

F=\frac{m (v_f -v_i)}{\Delta t}=\frac{(0.16 kg)(44 m/s-(-38 m/s))}{0.002 s}=6560 N

7 0
3 years ago
Explain the difference between weight and mass in terms of vectors and scalars
Jet001 [13]

Answer:

hatdog.

Explanation:

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3 years ago
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A 35 kg child goes down a playground that is inclined at an angle of 27.5 degrees above the horizontal. Find the acceleration if
tigry1 [53]

Answer:

4.16m/s²

Explanation:

According to Newtons second law;

\sum Fx = ma_x\\Fm - Ff = ma_x\\W sin\theta - \mu R cos \theta = ma_x\\mg sin\theta - \mu mg cos\theta = ma_x\\

Fm is the moving force

\mu is  the coefficient of kinetic friction between the child and the slide

m is the mass

g is the acceleration due to gravity

a is the acceleration of the child

Substitute the given values and get the acceleration as shown;

35(9.8)sin27.5 - 0.415(35)(cos27.5) = 35a

158.38-12.88 = 35a

145.49 = 35a

a = 145.49/35

a = 4.16m/s²

Hence the acceleration of the body is  4.16m/s²

4 0
3 years ago
If a cheetah goes from 10 m/s to 30 m/s in 5 seconds, what was her acceleration?
grandymaker [24]

Explanation:

Y CHETOS XDXDXX

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5 0
3 years ago
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