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3 days ago

If you double the mass of an object while leaving the net force unchanged what is the result

Physics

1 answer:

kicyunya [2.9K]3 days ago

5 0

Answer:

If the net force acting on an object doubles, its acceleration also doubles. Conversely, if the mass is doubled, the acceleration will be halved. If both the net force and mass are doubled, the acceleration remains the same.

Explanation:

[[TAG_9]][[TAG_10]]

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The radius of the earth's orbit around the sun (assumed to be circular) is 1.50 108 km, and the earth travels around this orbit

kicyunya [2911]

(a) 29,905 m/s

The speed of Earth's orbit around the Sun can be obtained by dividing the total distance of the orbit by the duration of the orbital period:

$v=\frac{2\pi r}{T}$

where

$r=1.50 \cdot 10^8 km = 1.50 \cdot 10^{11} m$ represents the radius of the orbit

$T=365 d \cdot 24 h/d \cdot 60 min/h \cdot 60 s/min =3.15 \cdot 10^7 s$ is the period taken for a complete orbit by Earth

By plugging values into this formula, we can calculate the orbital speed:

$v=\frac{2\pi (1.50\cdot 10^{11} m)}{(3.15\cdot 10^7 s)}=29,905 m/s$

(b) $5.96\cdot 10^{-3} m/s^2$

The centripetal acceleration that Earth experiences as it is drawn toward the sun is called radial acceleration:

$a=\frac{v^2}{r}$

where

v = 29,905 m/s is the orbital speed

$r=1.50 \cdot 10^8 km = 1.50 \cdot 10^{11} m$ is the radius of the orbit

By inserting into the relevant equation, we can derive:

$a=\frac{(29,905 m/s)^2}{(1.50\cdot 10^{11} m)}=5.96\cdot 10^{-3} m/s^2$

(c) 6,801 m/s

For the planet Uranus, we can determine:

$r=2.87 \cdot 10^9 km = 2.87 \cdot 10^{12} m$ is the radius of its orbit

$T=84.02 y \cdot 365 d/y \cdot 24 h/d \cdot 60 min/h \cdot 60 s/min =2.65 \cdot 10^9 s$ is the time it takes for one complete orbit

Thus, the orbital speed is calculated as:

$v=\frac{2\pi (2.87\cdot 10^{12} m)}{(2.65\cdot 10^9 s)}=6,801 m/s$

(d) $1.61\cdot 10^{-5} m/s^2$

For Uranus, with v = 6,801 m/s marking the orbital speed

$r=2.87 \cdot 10^9 km = 2.87 \cdot 10^{12} m$ is the orbital radius

Thus, the radial acceleration can be expressed as:

$a=\frac{(6,801 m/s)^2}{(2.87\cdot 10^{12} m)}=1.61\cdot 10^{-5} m/s^2$

6 0

1 month ago

Two experiments are performed on an object to determine how much the object resists a change in its state of motion while at res

inna [2732]

Answer:

The first experiment measures inertial mass, while the second experiment measures gravitational mass.

Explanation:

A student conducts two different experiments to observe resistance to changes in motion, both when at rest and in motion.

In the initial experiment, an object is forcefully pushed against a flat surface while its speed is tracked by a sensor. This setup involves work done against the object's inertia, identifying the mass as inertial mass.

Conversely, in the subsequent experiment, the object is lifted or thrown upward with an applied force and the speed is recorded. Here, the mass refers to gravitational mass, as the work performed combats gravity or the object's weight.

5 0

28 days ago

You are conducting an experiment inside a train car that may move along level rail tracks. A load is hung from the ceiling on a

serg [3228]

Answer:

The accurate statements are

2. The train is not an inertial frame of reference.

5. The train could be moving at a constant velocity in a circular path.

8. The train must be undergoing acceleration.

Explanation:

As we observe that the string forms an angle with the horizontal

we can formulate the force equation relevant to the given ball

$F_x = Tcos\theta$

$ma = Tcos45$

similarly in the Y direction

$mg = Tsin45$

Thus we conclude

$\frac{ma}{mg} = cot 45$

$a = g cot45$

This leads us to deduce that the train is accelerating with an acceleration identical to that of gravity

The correct statements will be

2. The train is not an inertial frame of reference.

5. The train could be moving at a constant speed in a circular path.

8. The train must be experiencing acceleration.

8 0

16 days ago

 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer

serg [3228]

Response:

a) The mug makes contact with the ground 0.7425m from the bar's end. b) |V|=5.08m/s θ= -72.82°

Clarification:

To address this issue, we begin with a diagram depicting the situation. (refer to the attached illustration).

The illustration shows that the problem involves motion in two dimensions. To determine how far from the bar the mug lands, we need to find the time the mug remains airborne by examining its vertical motion.

To compute the time, we utilize the following formula with the known values:

$y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}$

We have $y_{f}=0$ and $v_{y0}=0$ , allowing us to simplify the equation to:

$0=y_{0}+\frac{1}{2}at^{2}$

Now, we can calculate for t:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

We know $y_{0}=1.20m$ and $a=g=-9.8m/s^{2}$

The negative gravity indicates the downward motion of the mug. Hence, we substitute these values into the provided formula:

$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

Which results in:

t=0.495s

This time helps us evaluate the horizontal distance the mug traverses. Since:

$V_{x}=\frac{x}{t}$

Solving for x, we have:

$x=V_{x}t$

Substituting the known values yields:

x=(1.5m/s)(0.495s)

This calculates to:

x=0.7425m

b) With the time determining when the mug strikes the ground established, we can find the final velocity in the vertical direction using the formula:

$a=\frac{v_{f}-v_{0}}{t}$

The initial vertical velocity being zero simplifies our calculations:

$a=\frac{v_{f}}{t}$

Thus, we can determine the final velocity:

$V_{yf}=at$

Given that the acceleration equates to gravity (showing a downward effect), we substitute that alongside the previously found time:

$V_{yf}=(-9.8m/s^{2})(0.495s)$

This leads to:

$V_{yf}=-4.851m/s$

Now, we ascertain the velocity components:

$V_{xf}=1.5m/s$ and $V_{yf]=-4.851m/s$

Next, we find the speed by calculating the vector's magnitude:

$|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}$

<pThus:

$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

Yielding:

|V|=5.08m/s

Lastly, to ascertain the impact direction, we apply the equation:

$\theta = tan^{-1} (\frac{V_{y}}{V_{x}})$

<pFulfilling this provides:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

<pLeading to:

$\theta = -72.82^{o}$

4 0

1 month ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

inna [2732]

Answer:

$\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu$ ≈ 3077.34

Explanation:

To calculate the flux of F (vector field) across surface S, where

F(x,y,z) = y i − x j + $z^{2}$ k

and S(u,v) = u cos v i + u sin v j + v k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 5π

We will evaluate the following integral:

$\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu$

Substituting for the surface

x = u cos v

y = u sin v

z = v

Then

F(S(u,v)) = u sin v i - u cos v j + $v^{2}$ k

The normal vector N is computed as

$N = S_{u}XS_{v}$

Where:

$S_{u} = =$

$S_{v} = =$

N = < cos v, sin v, 0 > X <- u sin v, u cos v, 2v

N = < 2v sin v, -2v cos v, u >

F(S(u,v)).N = < u sin v, -u cos v, $v^{2}$ >. < 2v sin v, -2v cos v, u >

F(S(u,v)).N = 2uv + u $v^{2}$

Thus

$\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {2uv}+u v^{2}\,dvdu$ ≈ 3077.34

8 0

8 days ago