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2 years ago

13

a ball is rolled uphill a distance of 3 meters before it slows,stops,and begins to roll back. the ball rolls downhill 6 meters b

efore coming to rest against a tree. what is the magnitude of the balls displacement?

1 answer:

Leona [35]2 years ago

3 0

The answer to your question is 3 meters

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A 75.0-kg man standing on a bathroom scale in an elevator. Calculate the scale in N, reading if the elevator moves upward at a c

Salsk061 [2.6K]

The scale in N, reading if the elevator moves upward at a constant speed of 1.5 m/s^2 is 862.5 N.

weight of man = 75kg

speed of elevator, a = 1.5 $m/ s^{2}$

$F - w = ma \\$

$F = w + ma$

$F = m ( a +g )$

$F = 75 ( 1.5 + 10 ) \\$

$F = 75 ( 11.5 )$

$F = 862.5 N$

So, the scale reading in the elevator is greater than his 862.5 N weight. This indicates that the person is being propelled upward by the scale, which it must do in order to do so, with a force larger than his weight. According to what you experience in quickly accelerating or slowly moving elevators, it is obvious that the faster the elevator acceleration, the greater the scale reading.

Speed can be defines as the pace at which the position of an object changes in any direction. Since speed simply has a direction and no magnitude, it is a scalar quantity.

Learn more about speed here:-

brainly.com/question/19127881

#SPJ4

8 0

2 years ago

(a) An elevator of mass m moving upward has two forces acting on it: the upward force of tension in the cable and the downward f

Katen [24]

Answer: T is greater

Explanation:

Since the elevator is moving against gravity more work will be done on the rope

T= m(g+a)

8 0

2 years ago

Read 2 more answers

A 0.500-nm x-ray photon is deected through 134 in a Compton scattering event. At what angle (with respect to the incident beam)

natita [175]

Answer:

The angle of recoil electron with respect to incident beam of photon is 22.90°.

Explanation:

Compton Scattering is the process of scattering of X-rays by a charge particle like electron.

The angle of the recoiling electron with respect to the incident beam is determine by the relation :

$\cot\phi = (1+\frac{hf}{m_{e}c^{2} })\tan\frac{\theta }{2}$ ....(1)

Here ∅ is angle of recoil electron, θ is the scattered angle, h is Planck's constant, $m_{e}$ is mass of electron, c is speed of light and f is the frequency of the x-ray photon.

We know that, f = c/λ ......(2)

Here λ is wavelength of x-ray photon.

Rearrange equation (1) with the help of equation (1) in terms of λ .

$\cot\phi = (1+\frac{h}{m_{e}c\lambda })\tan\frac{\theta }{2}$

Substitute 6.6 x 10⁻³⁴ m² kg s⁻¹ for h, 9.1 x 10⁻³¹ kg for $m_{e}$ , 3 x 10⁸ m/s for c, 0.500 x 10⁻⁹ m for λ and 134° for θ in the above equation.

$\cot\phi = (1+\frac{6.6\times10^{-34} }{9.1\times10^{-31}\times3\times10^{8}\times0.5\times10^{-9} })\tan\frac{134 }{2}$

$\cot\phi=2.37$

$\phi$ = 22.90°

8 0

3 years ago

An unbalanced force gives a 2.00 kg mass an acceleration of 5.00 m/s? What is the force applied to the object?

valentinak56 [21]

Answer:

10N

Explanation:

Equation: ΣF = ma

Fapp = ma

Fapp = (2kg)(5m/s^2) (im guessing you mean 5.00 m/s^2 not m/s)

Fapp = 10*kg*m/s^2

Fapp = 10N

5 0

2 years ago

You place a light bulb 8 cm in front of a concave mirror. You then move a sheet of paper back and forth in front of the mirror.

Alika [10]

sorry - late reply...just stumbled across tis...hope u can still use it :)

By the mirror equation: 1/di + 1/do = 1/f

<span>
</span>

<span>where di = distance to image = +12cm (+ for real image)</span>

and do = distance to object = +8cm

Substitute and solve for f, the focal length

<span><span>
1/12 + 1/8 = 1/f
</span><span>
1/f = (8 + 12) / 12 * 8 = 20/96
</span><span>
so f = 96/20 = 4.8 cm</span>
</span>

5 0

3 years ago