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tamaranim1 [39]

3 years ago

6

Bill is throwing a football at four targets and attempting to knock them over. Which of the following targets will be hardest fo

r him to knock over? A. the 100-gram lead target. B. the 50-gram plastic target. C. the 150-gram nylon target. D. the 25-gram silicone target

1 answer:

Naddika [18.5K]3 years ago

7 0

100 g lead target. Lead has a very high density which means even a very small volume and weight quite a lot. So because the lead target weighs only 100 g it's going to be small in size compared to other targets which are all made of lesser dense materials than lead.

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What is the average velocity of a train moving along a straight track if its displacement is 192 m was during a time period of 8

Alex

Answer:

The average velocity of a train moving along a straight track if its displacement is 192 m was during a time period of 8.0 s is 24 $\frac{m}{s}$ .

Explanation:

Velocity is a physical quantity that expresses the relationship between the space traveled by an object and the time used for it. Then, the average velocity relates the change in position to the time taken to effect that change.

$velocity=\frac{displacement}{time}$

Velocity considers the direction in which an object moves, so it is considered a vector magnitude.

In this case, the displacement is 192 m and the time period is 8 s. Replacing:

$velocity=\frac{192 m}{8 s}$

Solving:

velocity= 24 $\frac{m}{s}$

<em><u>The average velocity of a train moving along a straight track if its displacement is 192 m was during a time period of 8.0 s is 24 </u></em> $\frac{m}{s}$ <em><u>.</u></em>

3 0

2 years ago

A 5.00 kg rock whose density is 4300 kg/m3 is suspended by a string such that half of the rock's volume is under water. You may

12345 [234]

Answer:

The tension on the string is $T = 43.302 \ N$

Explanation:

From the question we are told that

The mass of the rock is $m_r = 5.00 \ kg = 5000 \ g$

The density of the rock is $\rho = 4300 \ kg/m^3 = 4.3 g/dm^3$

Generally the volume of the rock is mathematically evaluated as

$V = \frac{m_r}{\rho}$

substituting values

$V = \frac{5000}{4.3}$

$V = 1162.7 \ dm^3$

The volume of the rock immersed in water is

$V_w = \frac{V}{2}$

substituting values

$V_w = \frac{1162.7 }{2}$

$V_w = 581.4 \ dm^3$

mass of water been displaced by the this volume is

$m_w = V_w$ According to Archimedes principle

=> $m_w = 581.4 \ g$

$m_w = 0.5814 \ kg$

The weight of the water displace is

$W _w = m_w * g$

$W _w = 0.5814 * 9.8$

$W _w = 5.698 \ N$

The actual weight of the rock is

$W_r = m_r * g$

$W_r = 5.0 * 9.8$

$W_r = 49.0 \ N$

The tension on the string is

$T = W_r - W_w$

substituting values

$T = 49.0 - 5.698$

$T = 43.302 \ N$

4 0

2 years ago

A student on an amusement park ride moves in circular path with a radius of 3.5 m once every 4.5 s. What is the tangential veloc

nata0808 [166]

Answer:

the tangential velocity of the student is 4.89 m/s.

Explanation:

Given;

the radius of the circular path, r = 3.5 m

duration of the motion, t = 4.5 s

let the student's tangential velocity = v

The tangential velocity of the student is calculated as follows;

$v = \frac{2\pi r}{t} \\\\v = \frac{2 \pi \times 3.5}{4.5} \\\\v = 4.89 \ m/s$

Therefore, the tangential velocity of the student is 4.89 m/s.

6 0

2 years ago

A bicycle wheel with radius 0.3 m rotates from rest to 3 rev/s in 5 s. What is the magnitude and direction of the total accelera

AlekseyPX

Answer:

Explanation:

Given

Radius of bicycle wheel $r=0.3\ m$

Initial angular velocity $\omega _0=0$

It rotates 3 revolution in 5 s therefore

$\omega =2\pi 3=\6\pi =18.85\ rad/s$

using $\omega =\omega _0+\alpha t$

where $\alpha =angular\ acceleration$

$\omega =Final\ angular\ velocity$

$t=time$

$\alpha =\frac{18.85}{5}=3.77 rad/s^2$

Total acceleration of any point will be a vector sum of tangential acceleration and centripetal acceleration

$\omega at t=1$

$\omega =0+3.77\times 1=3.77 rad/s$

$a_c=\omega ^2\cdot r$

$a_c=(3.77)^2\cdot 0.3=4.26 m/s^2$

Tangential acceleration $a_t=\alpha \times r$

$a_t=3.77\times 0.3=1.13 m/s^2$

$a_{net}=\sqrt{a_t^2+a_c^2}$

$a_{net}=\sqrt{(1.13)^2+(4.26)^2}$

$a_{net}=4.41 m/s^2$

7 0

3 years ago

How many stars are in a galaxy? hundreds thousands millions billions.

faust18 [17]

Answer:

Billions :)

<em>hope this helps! <3</em>

7 0

1 year ago