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NeX [460]
3 years ago
13

A ball was dropped and had a mass of .2 kg and was falling with a force of 2 N, what was its acceleration?

Physics
1 answer:
Musya8 [376]3 years ago
7 0

Answer:

10m/s I believe

Explanation:

F=MA

A=F/M

A=2/0.2

A=10m/s

You might be interested in
Three small masses are positioned as follows: 2.0 kg at (0.0 m, 0.0 m), 2.0 kg at (2.0 m, 0.0 m), and 4.0 kg at (2.0 m, 1.0 m).
melamori03 [73]
Refer to the diagram shown below.

The given data is

mass, kg   Coordinates. m
-------------   -----------------
   2               (0, 0)
   2               (2, 0)
   4               (2, 1)

Total mass, M = 2+2+4 = 8kg
Let (x,y) be the coordinates of M.

Then, taking moments about the origin, we obtain
8x = 2*0 + 2*2 + 4*2 = 12
x = 1.5 m

8y = 2*0 + 2*0 + 4*1 = 4
y = 0.5 m

Answer:  (1.5, 0.5) m




6 0
2 years ago
In 1999, Robbie Knievel was the first to jump the Grand Canyon on a motorcycle. At a narrow part of the canyon (65 m wide) and t
vfiekz [6]

Answer:

His launching angle was 14.72°

Explanation:

Please, see the figure for a graphic representation of the problem.

In a parabolic movement, the velocity and displacement vectors are two-component vectors because the object moves along the horizontal and vertical axis.

The horizontal component of the velocity is constant, while the vertical component has a negative acceleration due to gravity. Then, the velocity can be written as follows:

v = (vx, vy)

where vx is the component of v in the horizontal and vy is the component of v in the vertical.

In terms of the launch angle, each component of the initial velocity can be written using the trigonometric rules of a right triangle (see attached figure):

sin angle = opposite / hypotenuse

cos angle = adjacent / hypotenuse

In our case, the side opposite the angle is the module of v0y and the side adjacent to the angle is the module of vx. The hypotenuse is the module of the initial velocity (v0). Then:

sin angle = v0y / v0  then: v0y = v0 * sin angle

In the same way for vx:

vx = v0 * cos angle

Using the equation for velocity in the x-axis we can find the equation for the horizontal position:

dx / dt = v0 * cos angle

dx = (v0 * cos angle) dt (integrating from initial position, x0, to position at time t and from t = 0 and t = t)

x - x0 = v0 t cos angle

x = x0 + v0 t cos angle

For the displacement in the y-axis, the velocity is not constant because the acceleration of the gravity:

dvy / dt = g ( separating variables and integrating from v0y and vy and from t = 0 and t)

vy -v0y = g t

vy = v0y + g t

vy = v0 * sin angle + g t

The position will be:

dy/dt = v0 * sin angle + g t

dy = v0 sin angle dt + g t dt (integrating from y = y0 and y and from t = 0 and t)

y = y0 + v0 t sin angle + 1/2 g t²

The displacement vector at a time "t" will be:

r = (x0 + v0 t cos angle, y0 + v0 t sin angle + 1/2 g t²)

If the launching and landing positions are at the same height, then the displacement vector, when the object lands, will be (see figure)

r = (x0 + v0 t cos angle, 0)

The module of this vector will be the the total displacement (65 m)

module of r = \sqrt{(x0 + v0* t* cos angle)^{2} }  

65 m = x0 + v0 t cos angle ( x0 = 0)

65 m / v0 cos angle = t

Then, using the equation for the position in the y-axis:

y = y0 + v0 t sin angle + 1/2 g t²

0 =  y0 + v0 t sin angle + 1/2 g t²

replacing t =  65 m / v0 cos angle and y0 = 0

0 = 65m (v0 sin angle / v0 cos angle) + 1/2 g (65m / v0 cos angle)²  

cancelating v0:

0 = 65m (sin angle / cos angle) + 1/2 g * (65m)² / (v0² cos² angle)

-65m (sin angle / cos angle) = 1/2 g * (65m)² / (v0² cos² angle)  

using g = -9.8 m/s²

-(sin angle / cos angle) * (cos² angle) = -318.5 m²/ s² / v0²

sin angle * cos angle = 318.5 m²/ s² / (36 m/s)²

(using trigonometric identity: sin x cos x = sin (2x) / 2

sin (2* angle) /2 = 0.25

sin (2* angle) = 0.49

2 * angle = 29.44

<u>angle = 14.72°</u>

3 0
3 years ago
Las condiciones iniciales de un gas son 3000 cm3
slava [35]

Answer:

T'=92.70°C

Explanation:

To find the temperature of the gas you use the equation for ideal gases:

PV=nRT

V: volume = 3000cm^3 = 3L

P: pressure = 1250mmHg; 1 mmHg = 0.001315 atm

n: number of moles

R: ideal gas constant = 0.082 atm.L/mol.K

T: temperature = 27°C = 300.15K

For the given values you firs calculate the number n of moles:

n=\frac{PV}{RT}=\frac{(1520[0.001315atm])(3L)}{(0.082\frac{atm.L}{mol.K})(300.15K)}=0.200moles

this values of moles must conserve when the other parameter change. Hence, you have V'=2L and P'=3atm. The new temperature is given by:

T'=\frac{P'V'}{nR}=\frac{(3atm)(2L)}{(0.200\ moles)(0.082\frac{atm.L}{mol.K})}=365.85K=92.70\°C

hence, T'=92.70°C

8 0
3 years ago
1. Driving at slower speeds than traffic flow _____________ .
r-ruslan [8.4K]

<u>The correct option is (D). The strength of electric field depends on the amount of charge that produces the field as well as the distance from the charge. </u>

<u> </u>

Further Explanation:

The electric field intensity at a point is the measure of the force exerted by a charge particle on another charge particle in the particular area of its strength.

The electric field intensity at a distance d due to a static charge having charge q is directly proportional to the amount of charge and inversely proportional to the square of the distance between them.

The Electric field intensity due to a charge is given as:

E = \dfrac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}

Here, E is the electric field intensity, q is the amount of charge and r is the distance of the charge from the point.

The above expression of electric field shows that the electric field intensity at a point depends on the amount of charge as well as the distance of the point from the charge.

<u>Thus, the correct option is (D). The strength of electric field depends on the amount of charge that produces the field as well as the distance from the charge. </u>

<u> </u>

Learn More:

1. The results of Rutherford’s gold foil experiment demonstrated that thehttps://brainly.com/question/1542931

2. A capacitor with capacitance (c) = 4.50 μf is connected to a 12.0 v batteryhttps://brainly.com/question/8892837

3. A point charge of 7.45 pc is fixed at the originhttps://brainly.com/question/11098511

Answer Details:

Grade: Senior school

Subject: Physics

Chapter: Electrostatics

Keywords:  Strength, electric field, charge, distance, electric field intensity, magnitude of charge, electrostatic, test charge, kq/r^2.

7 0
3 years ago
Read 2 more answers
Compressions and rarefactions are characteristic of
Korvikt [17]

Answer:

Of longitudinal waves

Explanation:

Depending on the direction of the oscillation, there are two types of waves:

- Transverse waves: in a transverse wave, the oscillations occur perpendicularly to the direction of propagation of the wave. Examples are electromagnetic waves.

- Longitudinal waves: in a longitudinal wave, the oscillations occur parallel to the direction of propagation of the wave. In such a wave, the oscillations are produced by alternating regions of higher density of particles, called compressions, and regions of lower density of particles, called rarefactions. Examples of longitudinal waves are sound waves.

6 0
3 years ago
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