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3 years ago

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Question 7 Unsaved Brycen can cover half a basketball court (about 14 meters) in 4.0 seconds flat! How fast can he run? Question

7 options:
18 m/s

10 m/s

3.5 m/s

0.28 m/s

1 answer:

lana66690 [7]3 years ago

7 0

D = distance traveled = 14 m

t = time taken = 4 sec

v = speed of Brycen

the formula that relates speed, time and distance is given as

speed = distance/time

speed of Brycen is given as

v = D/t

inserting the values

v = 14/4

v = 3.5 m/s

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While a roofer is working on a roof that slants at 37.0 ∘∘ above the horizontal, he accidentally nudges his 92.0 NN toolbox, cau

eduard

Answer:

The speed is $v =8.17 m/s$

Explanation:

From the question we are told that

The angle of slant is $\theta = 37.0^o$

The weight of the toolbox is $W_t = 92.0N$

The mass of the toolbox is $m = \frac{92}{9.8} = 9.286kg$

The start point is $d = 4.25m$ from lower edge of roof

The kinetic frictional force is $F_f = 22.0N$

Generally the net work done on this tool box can be mathematically represented as

$Net \ work done = Workdone \ due \ to \ Weight + Workdone \ due \ to \ Friction$

The workdone due to weigh is = $mgsin \theta * d$

The workdone due to friction is = $F_f \ cos\theta * d$

Substituting this into the equation for net workdone

$W_{net} = mgsin\theta * d + F_f \ cos \theta *d$

Substituting values

$W_{net} = 92 * sin (37) * 4.25 + 22 cos (37) * 4.25$

$= 309.98 J$

According to work energy theorem

$W_{net} = \Delta Kinetic \ Energy$

$W_{net} = \frac{1}{2} m (v - u)^2$

From the question we are told that it started from rest so u = 0 m/s

$W_{net} = \frac{1}{2} * m v^2$

Making v the subject

$v = \sqrt{\frac{2 W_{net}}{m} }$

Substituting value

$v = \sqrt{\frac{2 * 309.98}{9.286} }$

$v =8.17 m/s$

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3 years ago

The density of mercury is 13.5 g/cm3

adoni [48]

Answer:

DO IY.T

Explanation:

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3 years ago

Height of cannon 5 m, initial speed of projectile 15m/s, angle of launch 0 degrees. What is the range and time in the air? Pleas

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Answer:

<em>The range is 15.15 m and the time in the air is 1.01 s</em>

Explanation:

<u>Horizontal Motion</u>

When an object is thrown horizontally (with angle 0°) with a speed v from a height h, it follows a curved path ruled exclusively by gravity until it eventually hits the ground.

The range or maximum horizontal distance traveled by the object can be calculated as follows:

$\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}$

To calculate the time the object takes to hit the ground, we use the equation below:

$\displaystyle t=\sqrt{\frac{2h}{g}}$

The cannon is shot from a height of h=5 m with an initial speed of v=15 m/s. The range is calculated below:

$\displaystyle d=15\cdot\sqrt{\frac {2*5}{9.8}}=15*1.01$

d = 15.15 m

The time in the air is:

$\displaystyle t=\sqrt{\frac{2*5}{9.8}}$

t = 1.01 s

The range is 15.15 m and the time in the air is 1.01 s

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Explanation:

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