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4 years ago

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An object is moving north with an initial velocity of 14 m/s accelerates 5m/s for 20 seconds. What is the final velocity of the

object

1 answer:

olga2289 [7]4 years ago

6 0

Use the kinematic equation: Vf=Vi+at

Then plug;

Vi=14 m/s

a=5 m/s²

t=20 s. Therefore;

Vf=14+(5*20)

Vf=114 m/s.

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Gravity.: Gravity is the force that acts at a right angle to the path of an orbiting object.

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4 years ago

Read 2 more answers

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

Which statement describes a force acting on an object?

Aleonysh [2.5K]

Answer: B. A magnet pulls a nail towards it

Explanation: Out of the list of answers, this is the only answer where a force is acting on an object. the act of the magnet pulling on the nail is the force, and the nail is the object.

7 0

4 years ago

Physics , help please

zaharov [31]

Acceleration in a velocity vs time graph is just the slope at that point. The reason for that is because the definition of acceleration is the change in velocity per unit of time. In this case we want instantaneous, which is the derivative or tangent line at that point.

At 3s we can see the slope is 0, so that means his acceleration is zero. That means he was moving at a constant velocity

At 5s we can see that the slope is negative. And from 5s to 6s the change in velocity is -5m/s^2

At 7s we can see the slope is very positive. And from 7s to 8s the change in velocity is +15m/s^2

And again, at 9s the slope is 0 so his acceleration is also zero. He’s moving at a constant velocity

If you take the integral of a velocity vs time graph, you get position. So the area underneath a velocity vs time graph is the distance traveled. Anything below the x axis is considered negative distance. We need to take the area of a triangle and the area of two rectangles to find the distance.

So, let’s do the two rectangles first. From 8s to 9s it is a width of 1 and a length of 40. So the area would be 40 meters. Let’s do the second rectangle. From 7s to 8s it is a width of 1. Then the length goes up to 25. So the area is 25 meters.

Now the triangle, the base is 1 and the height is 15. Divide 15 in half to get 7.5 meters

25 + 40 + 7.5 = 72.5 meters

6 0

3 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

4 years ago