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4 years ago

The reaction of two positively charged objects that are near each other

Physics

1 answer:

lilavasa [31]4 years ago

5 0

The objects will repel.

<u>Explanation</u>:

Charges are of two types - like and unlike. They are like charges if both are positive or if both are negative. They are unlike charges if one is positive and the other is negative.
When two objects of positive charge are near each other, they both have like charges.
When two like charges are near each other, they will repel each other.
When two dislike charges are near each other, they will attract each other.

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A person throws a stone from the corner edge of a building. The stone's initial velocity is 28.0 m/s directed at 43.0° above the

Naya [18.7K]

The stone's acceleration, velocity, and position vectors at time $t$ are

$\mathbf a(t)=-g\,\mathbf j$

$\mathbf v(t)=v_{i,x}\,\mathbf i+\left(v_{i,y}-gt\right)\,\mathbf j$

$\mathbf r(t)=v_{i,x}t\,\mathbf i+\left(y_i+v_{i,y}t-\dfrac g2t^2\right)\,\mathbf j$

where

$g=9.80\dfrac{\rm m}{\mathrm s^2}$

$v_{i,x}=\left(28.0\dfrac{\rm m}{\rm s}\right)\cos43.0^\circ\approx20.478\dfrac{\rm m}{\rm s}$

$v_{i,y}=\left(28.0\dfrac{\rm m}{\rm s}\right)\sin43.0^\circ\approx19.096\dfrac{\rm m}{\rm s}$

and $y_i$ is the height of the building and initial height of the rock.

(a) After 6.1 s, the stone has a height of 5 m. Set the vertical component $(\mathbf j)$ of the position vector to 5 m and solve for $y_i$ :

$5\,\mathrm m=y_i+\left(19.096\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.1\,\mathrm s)^2$

$\implies\boxed{y_i\approx70.8\,\mathrm m}$

(b) Evaluate the horizontal component $(\mathbf i)$ of the position vector when $t=6.1\,\mathrm s$ :

$\left(20.478\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)\approx\boxed{124.92\,\mathrm m}$

(c) The rock's velocity vector has a constant horizontal component, so that

$v_{f,x}=v_{i,x}\approx20.478\dfrac{\rm m}{\rm s}$

where $v_{f,x}$

For the vertical component, recall the formula,

${v_{f,y}}^2-{v_{i,y}}^2=2a\Delta y$

where $v_{i,y}$ and $v_{f,y}$ are the initial and final velocities, $a$ is the acceleration, and $\Delta y$ is the change in height.

When the rock hits the ground, it will have height $y_f=0$ . It's thrown from a height of $y_i$ , so $\Delta y=-y_i$ . The rock is effectively in freefall, so $a=-g$ . Solve for $v_{f,y}$ :

${v_{f,y}}^2-\left(19.096\dfrac{\rm m}{\rm s}\right)^2=2(-g)(-124.92\,\mathrm m)$

$\implies v_{f,y}\approx-53.039\dfrac{\rm m}{\rm s}$

(where we took the negative square root because we know that $v_{f,y}$ points in the downward direction)

So at the moment the rock hits the ground, its velocity vector is

$\mathbf v_f=\left(20.478\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(-53.039\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

which has a magnitude of

$\|\mathbf v_f\|=\sqrt{\left(20.478\dfrac{\rm m}{\rm s}\right)^2+\left(-53.039\dfrac{\rm m}{\rm s}\right)^2}\approx\boxed{56.855\dfrac{\rm m}{\rm s}}$

(d) The acceleration vector stays constant throughout, so

$\mathbf a(t)=\boxed{-g\,\mathbf j}$

4 0

3 years ago

Seesaw unbalanced force explain

wolverine [178]

Like a seesaw, it shows that the forces aren’t equal because if it was the seesaw would stay put

3 0

3 years ago

A mass of 10 g of oxygen fill a weighted pistonâ€“cylinder device at 20 kPa and 110Â°C. The device is now cooled until the tempe

mezya [45]

Answer:

The change of the volume of the device during this cooling is $14.3\times10^{-3}\ m^3$

Explanation:

Given that,

Mass of oxygen = 10 g

Pressure = 20 kPa

Initial temperature = 110°C

Final temperature = 0°C

We need to calculate the change of the volume of the device during this cooling

Using formula of change volume

$\Delta V=V_{2}-V_{1}$

$\Delta V=\dfrac{mR}{P}(T_{2}-T_{1})$

Put the value into the formula

$\Delta V=\dfrac{0.3125\times0.0821}{2.0265\times10^{9}}(383-273)$

$\Delta V=14.297\ L$

$\Delta V=14.3\times10^{-3}\ m^3$

Hence, The change of the volume of the device during this cooling is $14.3\times10^{-3}\ m^3$

6 0

3 years ago

If the torque required to loosen a nut that holds a wheel on a car has a magnitude of 55 n·m, what force must be exerted at the

erastova [34]

Either 175 N or 157 N depending upon how the value of 48Â° was measured from.
You didn't mention if the angle of 48Â° is from the lug wrench itself, or if it's from the normal to the lug wrench. So I'll solve for both cases and you'll need to select the desired answer.
Since we need a torque of 55 NÂ·m to loosen the nut and our lug wrench is 0.47 m long, that means that we need 55 NÂ·m / 0.47 m = 117 N of usefully applied force in order to loosen the nut. This figure will be used for both possible angles.
Ideally, the force will have a 0Â° degree difference from the normal and 100% of the force will be usefully applied. Any value greater than 0Â° will have the exerted force reduced by the cosine of the angle from the normal. Hence the term "cosine loss".
If the angle of 48Â° is from the normal to the lug wrench, the usefully applied power will be:
U = F*cos(48)
where
U = Useful force
F = Force applied
So solving for F and calculating gives:
U = F*cos(48)
U/cos(48) = F
117 N/0.669130606 = F
174.8537563 N = F
So 175 Newtons of force is required in this situation.
If the 48Â° is from the lug wrench itself, that means that the force is 90Â° - 48Â° = 42Â° from the normal. So doing the calculation again (this time from where we started plugging in values) we get
U/cos(42) = F
117/0.743144825 = F
157.4390294 = F
Or 157 Newtons is required for this case.

6 0

3 years ago

A friend asks you how much pressure is in your car tires. You know that the tire manufacturer recommends 30 psi, but it's been a

Rama09 [41]

Answer:

21 psi

Explanation:

The weight of the car is:

W = mg

W = 1000 kg * 9.8 m/s²

W = 9800 N

Divided by 4 tires, each tire supports:

F = W/4

F = 9800 N / 4

F = 2450 N

Pressure is force divided by area, so:

P = F / A

P = (2450 N) / (0.13 m × 0.13 m)

P ≈ 145,000 Pa

101,325 Pa is the same as 14.7 psi, so:

P ≈ 145,000 Pa × (14.7 psi / 101,325 Pa)

P ≈ 21 psi

3 0

3 years ago