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3 years ago

NEED THIS ANSWERED ASAP!!!! IN MIDDLE OF TEST RN HELP PLZ

2 answers:

4 0

The answer should be A, it does not exhibit projectile motion and follows a straight path down the ramp.

It's not in the air or anything so we definitely know it's not a projectile :D.

Marina86 [1]3 years ago

3 0

The correct answer for this question would be option A. When Barry is conducting an experiment and rolls a tennis ball down a ramp, the statement that best describes the motion of the tennis ball is that, it does not exhibit projectile motion and follows a straight path down the ramp. Hope this helps.

You might be interested in

A ball is thrown horizontally from the top of a 60 m building and lands 100 m from the base of the building. How long is the bal

zhannawk [14.2K]

Answer:

The ball is in the air for 3.5 seconds

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

Explanation:

A ball is thrown horizontally

That means the vertical component of the initial velocity $u_{y}=0$

The initial velocity is the horizontal component $u_{x}$

The ball is thrown from the top of a 60 m

That means the vertical displacement component y = 60 m

→ y = $u_{y}$ t + $\frac{1}{2}$ gt²

where g is the acceleration of gravity and t is the time

y = -60 m , g = -9.8 m/s² , $u_{y}=0$

Substitute these values in the rule

→ -60 = 0 + $\frac{1}{2}$ (-9.8)t²

→ -60 = -4.9t²

Divide both sides by -4.9

→ 12.2449 = t²

Take √ for both sides

∴ t = 3.5 seconds

* The ball is in the air for 3.5 seconds 

The initial velocity is the horizontal component $u_{x}$

The ball lands 100 meter from the base of the building

That means the horizontal displacement x = 100 m

→ x = $u_{x}$ t

→ t = 3.5 s , x = 100 m

Substitute these values in the rule

→ 100 = $u_{x}$ (3.5)

Divide both sides by 3.5

→ $u_{x}$ = 28.57 m/s

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is $v_{y}$

→ $v_{y}$ = $u_{y}$ + gt

→ $u_{y}$ = 0 , g = -9.8 m/s² , t = 3.5 s

Substitute these values in the rule

→ $v_{y}$ = 0 + (-9.8)(3.5)

→ $v_{y}$ = -34.3 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity v is the resultant vector of $v_{x}$ and $v_{y}$

→ Its magnetude is $v=\sqrt{(v_{x})^{2}+(v_{y})^{2}}$

→ Its direction $tan^{-1}\frac{v_{y}}{v_{x}}$

→ $v_{y}$ = 28.6 , $v_{y}$ = -34.3

Substitute this values in the rules above

→ $v=\sqrt{(28.6)^{2}+(-34.3)^{2}}=44.66$

→ Its direction $tan^{-1}\frac{-34.3}{28.6}=-50.18$

The negative sign means the direction is below the horizontal

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

7 0

2 years ago

Explain the relationship between speed, velocity and acceleration

Montano1993 [528]

Speed is the rate at which something covers a distance; velocity is the same but it takes into account whether it goes forwards or backwards; and acceleration is the rate of an increase in speed.

8 0

3 years ago

A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0° above the horizontal. The fire-f

zysi [14]

Answer:

8.8 m and 52.5 m

Explanation:

The vertical component and horizontal component of water velocity leaving the hose are

$v_v = vsin(\alpha) = 25sin(53^0) = 25*0.8 = 19.97 m/s$

$v_h = vcos(\alpha) = 25cos(53^0) = 25*0.6 = 15 m/s$

Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level

$s = v_vt + gt^2/2$

$10 = 19.97t - 9.8t^2/2$

$4.9t^2 - 19.97t + 10 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{19.9658877511823\pm \sqrt{(-19.9658877511823)^2 - 4*(4.9)*(10)}}{2*(4.9)}$

$t= \frac{19.9658877511823\pm14.24}{9.8}$

t = 3.49 or t = 0.58

We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down

t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building

$s_1 = v_ht_1 = 15*0.58 = 8.8 m$

$s_2 = v_ht_2 = 15*3.49 = 52.5m$

8 0

2 years ago

In September 2016 you purchased a 125 shares of Grainger stock for $224.84. It is now September 2019 and Grainger stock is curre

Temka [501]

Answer:

Gain in capital = $ 70.72

Explanation:

Given:

- The price of stocks when purchased P_o = $ 224.84

- The price of stocks when sold P_s = $ 295.56

Find:

what would be your capital gain (loss) on the sale, ignoring commissions

Solution:

- The capital gain or loss on the selling of stocks stems from the difference of buying and selling value of stocks. The original price of stock was P_o and the selling price would be P_s. The difference would be:

capital gain = P_s - P_o

capital gain = $295.56 - $224.84

capital gain = $ 70.72

- Hence, there would be a gain in capital if sold today for about $ 70.72.

6 0

3 years ago

g Initially, the motorcycle travels along a straight road with a speed of 35 m/s (this is almost 80 mph). The maximum decelerati

astraxan [27]

Given:

Initial speed of the motorcycle (u) = 35 m/s

Final speed of the motorcycle (v) = 0 m/s (Complete Stop)

Maximum deceleration of the motorcycle (a) = -1.2 m/s²

Required Equation:

$\boxed{\bf{ v = u + at}}$

Answer:

By substituting values in the equation, we get:

$\rm \longrightarrow 0 = 35 + ( - 1.2)t \\ \\ \rm \longrightarrow 0 = 35 - 1.2t \\ \\ \rm \longrightarrow 35 - 1.2t = 0 \\ \\ \rm \longrightarrow 35- 35 - 1.2t = 0 - 35 \\ \\ \rm \longrightarrow - 1.2t = - 35 \\ \\ \rm \longrightarrow \dfrac{ - 1.2t}{ - 1.2} = \dfrac{ - 35}{ - 1.2} \\ \\ \rm \longrightarrow t = 29.167 \: s$

$\therefore$ Time taken by motorcycle to come to a complete stop (t) = 29.167 s

4 0

3 years ago