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zmey [24]
3 years ago
11

A snowboarder goes down the hill with a slope of 28° if friction acts on him as he slides down which of the following is the cor

rect free body diagram for this situation

Physics
1 answer:
xxMikexx [17]3 years ago
6 0

Answer:

A

Explanation:

All of the frictions are the same, but weight always goes straight down so it can only be A or B. Since they are going down a slope, then the normal force must be sloped. A is the only one out of A and B with a sloped normal force, so it has to be A

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An underground cannon launches a cannonball from ground level at a 35 degree angle. the cannonball is shot with an initial veloc
user100 [1]

Answer:

Time = 1.75[s]; Distance traveled = 21.5 [m]; Max height = 15 [m]

Explanation:

First, we have to break down the velocity vector into the X & y components.

(v_{x})_{0} = 15 * cos( 35)= 12.28[m/s]\\(v_{y})_{0} = 15 * sin( 35)= 8.6[m/s]\\\\

To find the time t that lasts the ball of cannon in the air we must use the following equation of kinematics, in this equation the value of y is equal to zero because it will be proposed that the ball lands at the same level that was fired.

y=(v_{y} )_{0}-\frac{1}{2}*g*t^{2}   \\where:\\g=9.81[m/s^2]\\t = time[s]\\y=0[m]

0=8.6*t-\frac{1}{2}*9.81*t^{2}  \\4.905*t^{2}=8.6*t\\ t=1.75[s]

In order to find the distance traveled horizontally from the cannonball, we must use the speed kinematics equation in the X coordinate.

x = (v_{x})_{0}  *t\\x=12.28*1.75\\x=21.5 [m]

In order to find the last value, we must bear in mind that when the cannonball reaches the maximum height, the velocity in the component y is equal to zero, and we can find the value of and with the following kinematic equation

y = (v_{y})_{0} *t+\frac{1}{2} *g*(t)^{2} \\y = 0*t+\frac{1}{2} *9.81*(1.75)^{2}\\ y=15 [m]

6 0
3 years ago
1. A bicycle initially moving with a velocity
ki77a [65]

Answer:

\boxed {\boxed {\sf 15 \ m/s \ or \ 15 \ m*s^{-1}}}

Explanation:

We are asked to find the final velocity. We are given the acceleration, time, and initial velocity, so we can use the following kinematics formula.

v_f= v_i+ at

In this formula, v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and t is the time.

The bicycle has an initial velocity of 5.0 m *s⁻¹ or m/s, acceleration of 2 m/s², and a time of 5 seconds.

\bullet \  v_i = 5.0 \ m/s \\\bullet \  a= 2\ m/s^2\\\bullet \  t= 5  \ s

Substitute the values into the formula.

v_f=5.0 \ m/s + ( 2\ m/s^2 * 5 \ s)

Solve inside the parentheses.

  • \frac {2 \ m}{s^2}* 5 \ s = \frac{ 2 \ m}{s} * 5 = \frac{ 10 \ m}{s} = 10 \ m/s

v_f= 5.0 \ m/s + (10 \ m/s)

Add.

v_f= 15 \ m/s

The units can also be written as:

v_f= 15 \ m*s^{-1}

The bicycle's final velocity is 15 meters per second.

7 0
2 years ago
Devon places sliced potatoes in a package of aluminum foil. He puts a metal grate over a campfire and sets the package on the gr
sp2606 [1]

when he set the potatoes in the aluminum foil over the campfire because the heat was directly touching them and heating them

7 0
3 years ago
Read 2 more answers
Describe a technology used in space exploration.
Alex777 [14]

Answer:

High speed optical communication technology

To be able to communicate from the space to the earth and from earth to space is one of the most essential features required during space exploration.

Explanation:

Space exploration involves going into the space, beyond the earth's atmosphere. Landing on other planets and studying their details, going into deeper space beyond the planets to discover new cosmic events or structures is all a part of space exploration.

The key to analyse the studies and observations is being able to communicate the data collected, photos taken etc to the launch centers or space centers on earth. The space centers on earth should also be able to communicate with the persons or the satellites in space.

This is made possible using the optical communication technology which involves the use of optical fibers, lasers etc, since high speeds are more efficient during communication

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3 years ago
A leopard with a mass of 55.00 kg climbs 12.0 m up a tree. What is it’s gain in GPE
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The answer is: 6,468 j
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