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aleksandrvk [35]

4 years ago

10

A 5.5 kg bowling ball initially at rest is dropped from the top of a 12 m building. It hits the ground 1.75 s later. Find the ne

t external force on the falling ball.

1 answer:

Thepotemich [5.8K]4 years ago

7 0

We are given that:

m = 5.5 kg

d = 12 m

t = 1.75 s

Using the distance formula, we can get the acceleration of the ball:

d = (1/2) a*t^2

Rearranging:

a = 2d/t^2 = 2*(12 m)/(1.75 s)^2

a = 7.837 m/s^2

<span>Using Newton’s 2nd law, force is defines as:</span>

F = ma

F = 5.5 kg (7.837 m/s^2)

<span>F = 43.10 N</span>

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A constant force = 2.00 n + 3.00 n acts on a 5.00 kg object as it moves in a straight line from the position1 = 1.00 m + 1.00 m

nata0808 [166]

force acting on the object is given as

$F = 2.00\hat i + 3.00 \hat j$

initial and final positions are given as

$r_i = 1.00 \hat i + 1.00\hat j$

$r_f = 4.00 \hat i - 1.00\hat j$

now the displacement is given as

$d = r_f - r_i$

$d = (4.00 - 1.00)\hat i + (-1.00 - 1.00)\hat j$

$d = 3 \hat i - 2\hat j$

now the work done is given as

$W = F.d$

$W = (2.00\hat i + 3.00\hat j).(3\hat i - 2\hat j)$

$W = 0$

so there is no work done

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3 years ago

What was Galleos contribution to the study of motion ?

Rashid [163]

Answer:

B ) He was the first to systematically study force and motion.

Explanation:

I seen this one before and know the answer since there's 4 options

6 0

3 years ago

A tow rope pulls a 1450 kg truck, giving it an acceleration 1.25 m/s^2. What force does the rope exert?

bagirrra123 [75]

force=mass × acceleration

mass=force ÷ acceleration

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4 0

3 years ago

Electrons are continuously being knocked out of air molecules by cosmic ray particles from space. Once the electrons are free, t

nata0808 [166]

(a) $1.25\cdot 10^{-14} J$

The change in potential energy of the electron is given by:

$\Delta U=q E d$

where

$q=1.6\cdot 10^{-19}C$ is the magnitude of the electron's charge

$E=150 N/C$ is the magnitude of the electric field

d = 520 m is the distance through which the electron has moved

Substituting into the equation, we find

$\Delta U=(1.6\cdot 10^{-19}C)(150 N/C)(520 m)=1.25\cdot 10^{-14} J$

(b) 78 kV

The potential difference the electron has moved through is given by

$\Delta V=Ed$

where

$E=150 N/C$ is the magnitude of the electric field

d = 520 m is the distance through which the electron has moved

Substituting into the equation, we find

$\Delta V=Ed=(150 N/C)(520 m)=78,000 V=78 kV$

4 0

3 years ago

nordsb [41]

A) 320 count/min

B) 40 count/min

C) 80 count/min, 11400 years

Explanation:

A)

The activity of a radioactive sample is the number of decays per second in the sample.

The activity of a sample is therefore directly proportional to the number of nuclei in the sample:

$A\propto N$

where A is the activity and N the number of nuclei.

As a consequence, since the number of nuclei is proportional to the mass of the sample, the activity is also directly proportional to the mass of the sample:

$A\propto m$

where m is the mass of the sample.

In this problem:

- When the mass is $m_1 = 1 g$ , the activity is $A_1=16 count/min$

- When the mass is $m_2=20 g$ , the activity is $A_2$

So we can find A2 by using the rule of three:

$\frac{A_1}{m_1}=\frac{A_2}{m_2}\\A_2=A_1 \frac{m_2}{m_1}=(16)\frac{20}{1}=320 count/min$

B)

The equation describing the activity of a radioactive sample as a function of time is:

$A(t)= A_0 e^{-\lambda t}$ (1)

where

$A_0$ is the initial activity at time t = 0

t is the time

$\lambda$ is the decay constant, which gives the probability of decay

The decay constant can be found using the equation

$\lambda = \frac{ln2}{t_{1/2}}$

where $t_{1/2}$ is the half-life, which is the amount of time it takes for the radioactive sample to halve its activity.

In this problem, carbon-14 has half-life of

$t_{1/2}=5700 y$

So its decay constant is

$\lambda=\frac{ln2}{5700}=1.22\cdot 10^{-4} y^{-1}$

We also know that the tree died

t = 17,100 years ago

and that the initial activity was

$A_0 = 320 count/min$ (value calculated in part A, corresponding to a mass of 20 g)

So, substituting into eq(1), we find the new activity:

$A(17,100) = (320)e^{-(1.22\cdot 10^{-4})(17,100)}=40 count/min$

C)

We know that a sample of living wood has an activity of

$A=16 count/min$ per 1 g of mass.

Here we have 5 g of mass, therefore the activity of the sample when it was living was:

$A_0 = A\cdot 5 = (16)(5)=80 count/min$

Moreover, here we have a sample of 5 g, with current activity of $A=20 count/min$ : it means that its activity per gram of mass is

$A'=\frac{20}{5}=4 count/min$

We know that the activity halves after every half-life: Here the activity has became 1/4 of the original value, this means that 2 half-lives have passed, because:

- After 1 half-life, the activity drops from 16 count/min to 8 count/min

- After 2 half-lives, the activity dropd to 4 count/min

So the age of the wood is equal to 2 half-lives, which is:

$t=2t_{1/2}=2(5700)=11,400 y$

3 0

4 years ago