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2 years ago

15

A projectile is launched with a horizontal velocity of 20 m/s and an initial vertical velocity of 20 m/s. What is the projectile

's acceleration in the Horizontal direction? Verticle direction?

1 answer:

cestrela7 [59]2 years ago

6 0

Answer:

Vertical acceleration 9.8 m/s² downward

Horizontal acceleration 0.0 m/s²

assuming no air resistance.

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One student did an experiment on the rock cycle.

Nonamiya [84]

First, when the student added the layers of wax over each other, this became a representation of sedimentary rocks.

Then the student folded his/her palm and squeezed the layers of wax. This means that the student applied heat and pressure on the wax (sedimentary rocks)

Referring to the diagram below which represents the rock cycle, we will find that applying heat and pressure on sedimentary rocks would convert these rocks into metamorphic rocks.

Based on the above, the best choice would be:
<span>d. Heat and pressure can change sedimentary rocks into metamorphic rocks.</span>

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3 years ago

escribe the differences between the terrestrial and jovian planets and classify each of the 8 planets to the proper type

Radda [10]

Answer:

*******

Explanation:

is obviously the correct answer they are both so different yet so alike

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3 years ago

Read 2 more answers

A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a dra

bija089 [108]

Explanation:

Terminal velocity is given by:

$v_t=\sqrt{\frac{2mg}{\rho C_dA}}$

Here, m is the mass of the falling object, g is the gravitational acceleration, $C_d$ is the drag coefficient, $\rho$ is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

$A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg$

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is $1.28\frac{kg}{m^3}$ .

$v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}$

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

$v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}$

7 0

2 years ago

If the net force acting on an object is 0 N, you can be sure thr forces acting on the object are,

Bas_tet [7]

Answer:

A. Balanced

Explanation:

7 0

2 years ago

A comet is cruising through the solar system at a speed of 50,000 kilometers per hour foe 4 hours time. What is the total distan

Triss [41]

Answer:

<em>The distance covered by comet is </em> $200,000 km$

Explanation:

Speed is defined as the rate of change of distance with time. It is given by the equation speed= $\bold{\frac{distance}{time}}$

Thus distance= $speed*time$

In this problem it is given that speed of comet= $\frac{50,000km}{hr}$

time travelled by the comet= 4 hours

Thus distance= $speed*time$

= $500000*4$

= $\bold{200,000km}$

5 0

2 years ago