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3 years ago

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An apple suddenly drops a distance of 3.2 m from a tree. If the acceleration due to gravity is 9.8 m/s2, how long does it take t

he apple to reach the ground?. Give your answer to two decimal places.

1 answer:

mart [117]3 years ago

7 0

Answer:

0.33 seconds

Explanation:

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Roots that grow horizontally are called

Romashka [77]

I believe they are called Rhizomes.

8 0

3 years ago

Read 2 more answers

(1 point) A pilot flies in a straight path for 1 hour and 30 min. She then makes a course correction, heading 10 degrees to the

kati45 [8]

Answer:

The pilot is 2214.22 miles from her starting position

Explanation:

Since the pilot is traveling at a constant speed of 635 mph, the total distance traveled can be easily found as follows:

$1stleg: d_{1} = 1.5*635 = 952.5 mi\\2ndleg: d_{2} = 2*635 = 1270 mi$

There was a 10 degrees deviation, so the angle between the trajectory of both legs is 170 degrees.

The distance we need to find is that from the start of the first leg to the end of the second leg, those three distances form a triangle and since the side we're interested in is opposite to the 170 degrees angle, we can determine its length by the law of cosines:

$d_{3}^{2} =d_{1}^{2} + d_{2}^{2} -2*d_{1}*d_{2}*cos(170)\\d_{3}^{2} =952.5^{2} + 1270^{2} -2*952.5*1270*cos(170)\\d_{3}=2214.22mi$

The pilot is 2214.22 miles from her starting position

8 0

3 years ago

During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 4.02 s, how high above the point where i

alukav5142 [94]

Answer:

d = 19.796m

Explanation:

Since the ball is in the air for 4.02 seconds, the ball should reach the maximum point from the ground in half the total time, therefore, t=2.01s to reach maximum height. At the maximum height, the velocity in the y-direction is 0.

So we know t=2.01, vi=0, g=a=9.8m/s and we are solving for d.

Next, you look for a kinematic equation that has these parameters and the one you should choose is:

$d=vt+\frac{1}{2}at^2$

Now by substituting values in, we get

$d=(0\frac{m}{s})*(2.01s)+\frac{1}{2}(9.8\frac{m}{s^2})(2.01)^2$

d = 19.796m

7 0

3 years ago

The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge

lesya [120]

Answer:

Loss, $\Delta E=-10.63\ J$

Explanation:

Given that,

Mass of particle 1, $m_1=m =0.66\ kg$

Mass of particle 2, $m_2=7.4m =4.884\ kg$

Speed of particle 1, $v_1=2v_o=2\times 6=12\ m/s$

Speed of particle 2, $v_2=v_o=6\ m/s$

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

$m_1v_1+m_2v_2=(m_1+m_2)V$

$V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}$

$V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}$

V = 6.71 m/s

Initial kinetic energy before the collision,

$K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)$

$K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)$

$K_i=135.43\ J$

Final kinetic energy after the collision,

$K_f=\dfrac{1}{2}(m_1+m_2)V^2$

$K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2$

$K_f=124.80\ J$

Lost in kinetic energy,

$\Delta K=K_f-K_i$

$\Delta K=124.80-135.43$

$\Delta E=-10.63\ J$

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

7 0

3 years ago

A laboratory electromagnet produces a magnetic field of magnitude 1.38 T. A proton moves through this field with a speed of 5.86

Vladimir [108]

.Answer;

Using Fmax=qVB

F=(1.6*10^-19 C)(5.860*10^6 m/s)(1.38 T)

ANS=1.29*10^-12 N

2. Using Amax=Fmax/ m

Amax =(1.29*10^-12 N) / (1.67*10^-27 kg)

ANS=1.93*10^15 m/s^2*

3. No, the acceleration wouldn't be the same. Since The magnitude of the electron is equal to that of the proton, but the direction would be in the opposite direction and also Since an electron has a smaller mass than a proton

3 0

3 years ago