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rusak2 [61]
3 years ago
14

In an action movie, a stuntman leaps from the top of one building to the top of another building 5.2 m away. After a running sta

rt, he leaps at a velocity of 6.0 m/s at an angle of 15° above horizontal. Will he make it to the other roof, which is 2.9 m shorter than the building he jumps from?
Physics
1 answer:
Ksenya-84 [330]3 years ago
8 0
<span>stuntman's vertical velocity is vy= 6 sin 15=1.55m/s. Height that he goes up is, vertical kinetic energy = mgh 1/2 v² = gh, h=v²/(2g)=0.12 m. time wanted to go up= vy/9.8=0.16s, Time to fall through a height 0.12+2.9=3.02m is t=sqrt(3.02*2/9.8)=0.78 s Total time needed to go up and down is 0.78+0.16=0.94 s. to calculate the horizontal range, Horizontal velocity = 6 cos 15=5.8 m/s. Distance which he can cover is 5.8*0.94=5.44 m. If the distance between the two building is less than 5.44 m then he will be safe and he can jump that distance.</span>
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May you help me answer this​
Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is 2.2 m/s^2

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

1)

There are three Kepler's law of planetary motion:

  1. 1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
  2. 2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
  3. 3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, T^2 \propto r^3, where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

M g sin \theta - T = Ma (1)

where

Mg sin \theta is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

g=9.8 m/s^2 (acceleration of gravity)

\theta=35^{\circ}

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

T-mg sin \theta = ma (2)

where

m = 3.5 kg (mass of the block)

\theta=35^{\circ}

From (2) we get

T=mg sin \theta + ma

Substituting into (1),

M g sin \theta - mg sin \theta - ma = Ma

Solving for a,

a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2

2b)

The tension in the string can be calculated using the equation

T=mg sin \theta + ma

where

m = 3.5 kg (mass of lighter block)

g=9.8 m/s^2

\theta=35^{\circ}

a=2.2 m/s^2 (acceleration found in part 2)

Substituting,

T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

U=mgh

where

m is the mass of the object

g is the strength of the gravitational field

h is the heigth of the object relative to the ground

3b)

The kinetic energy of an object is given by

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

For the object in this problem,

m = 500 g = 0.5 kg

v = 3 m/s

Substituting, we find its kinetic energy:

K=\frac{1}{2}(0.5)(3)^2=2.25 J

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

And about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

7 0
3 years ago
Find the voltage across the 15 Ω resistor. .<br> 35 Ω<br> 20 Ω<br> 15 Ω<br> 10 V
almond37 [142]

Answer:

35 Ω

Explanation:

3 0
2 years ago
A tow truck is pulling 25,000 kg van with a tow hook that meets the van at an angle of 30° with the ground. How much force must
Law Incorporation [45]

Answer:

86605.08 N

Explanation:

The equation to calculate the force is:

Force = mass * acceleration

The force and the acceleration does not have the same direction in this case, so we need to decompose the force into its horizontal component, which is the force that will generate the horizontal acceleration:

Force_x = Force * cos(30)

Then, we have that:

Force_x = mass * acceleration

Force * cos(30) = 25000 * 3

Force * 0.866 = 75000

Force = 75000 / 0.866 = 86605.08 N

3 0
3 years ago
How do you find the net force acting on an object?
weqwewe [10]

Answer:

C. Add all the force vectors

Explanation:

The net force acting on an object is the vector sum of all the  forces on the object.

Remember, Newton's first law tells us a body at rest will remain at rest or that in uniform motion will continue in motion unless acted by unbalanced forces.These unbalanced forces act in all direction towards the body thus to get the net force you require a summation of all these force with respect to their magnitudes and directions.

For example a force of 3N towards the East direction acting on a body and another force of 2N towards the West direction on the same body will generate a net force of 1N towards the East direction.

4 0
3 years ago
A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 20.0 m/s. Then the truck trav
aniked [119]
<span>You can use the equation
V_xf = V_xi + a_x(t)

V_xf = 20.0m/s
V_xi = 0m/s
ax = 2.0 t

Thus, solve for t and get 10seconds and then take 5 seconds to break after 20 seconds of driving so for

a) 10 + 20 + 5 = 35 seconds

</span><span>for part b)
You can use the formula
Delta x/Delta t = average velocity
 
Need to find xf, knowing xi = 0

Thus, use the formula
 x_f = x_i + V_xi(t) + (1/2)a_x(t)^(2)
x_f = 0 + 0(10) + (1/2)(2.0)(10)^(2)
 x_f = 100m
 
so for the first 10 seconds the truck traveled 100ms At a speed of 20m/s

20m/s = xm/20s 20*20 = x
x = 400
 
thus we have 100+400 = 500m then it slows down from 500m to x_f
 
thus I use the equation
x_f = x_i + (1/2)(V_xf + V_xi)t
x_f = 500 + (1/2)(0 + 20)(5) x_f = 500 + 50
x_f = 550
 
therefore the total distance traveled is 550m
</span>
<span>to calculate average velocity
550/35 = 16m/s

thus V_xavg = 16m/s</span>
8 0
3 years ago
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