Please help me answer this

A horizontal line above the time axis of a speed vs. time graph means an object is ___.

miv72 [106K]

A horizontal line on a speed/time graph means a constant speed.

6 0

3 years ago

The system below has a friction force of 25 N acting on the cart which 8 kg. The mass hanging off the edge has a mass of 6 kg. F

photoshop1234 [79]

The cart will be pulled to the right by the hanging mass, so by Newton's second law, the net force on the cart is

T - 25 N = (8 kg) a

where T is the tension in the rope and a is the acceleration.

The hanging mass has a net force of

(6 kg) g - T = (6 kg) a

where g = 9.8 m/s².

Adding these equations together eliminates T, and we can solve for a :

(T - 25 N) + ((6 kg) g - T ) = (14 kg) a

33.8 N = (14 kg) a

a = (33.8 N) / (14 kg) ≈ 2.4 m/s²

Then the tension in the rope is

T - 25 N = (8 kg) (2.4 m/s²)

T ≈ 25 N + 19.31 N ≈ 44 N

5 0

2 years ago

17.

Zolol [24]

Answer:

I don't know this answer at all

Explanation:

I don't know about these problems

6 0

2 years ago

What happens to the force between the spheres when you increase the mass of one of the spheres?

andrew11 [14]

The force between the spheres increases when the mass increases in one of the spheres.

Explanation:

Newton law of universal gravity extends gravity beyond the earth's surface. This gravity depends directly on the mass of both objects and is inversely proportional to square of distance between their centers.

$\bold{F=\frac{G \times\left(m_{1} \times m_{2}\right)}{\left(r^{2}\right)}}$

Since gravity is directly proportional to “mass of both interacting objects”, stronger objects with greater gravitational force attract. If the mass of one object increases, gravity between them also increases. For example, if an object's mass of one double, force between them also doubles.

4 0

2 years ago

Read 2 more answers

A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribut

blsea [12.9K]

Answer:

1. $E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. $E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3.The results from part 1 and 2 agree when r = R.

Explanation:

The volume charge density is given as

$\rho (r) = \alpha (1-\frac{r}{R})$

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.

1. Since the cylinder is very long, Gauss’ Law can be applied.

$\int {\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

$\int\, da = 2\pi r h$

where ‘h’ is the length of the imaginary Gaussian surface.

$Q_{enc} = \int\limits^r_0 {\rho(r)h} \, dr = \alpha h \int\limits^r_0 {(1-r/R)} \, dr = \alpha h (r - \frac{r^2}{2R})\left \{ {{r=r} \atop {r=0}} \right. = \alpha h (\frac{2Rr - r^2}{2R})\\E2\pi rh = \alpha h \frac{2Rr - r^2}{2R\epsilon_0}\\E(r) = \alpha \frac{2R - r}{4\pi \epsilon_0 R}\\E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

$Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = \alpha \int\limits^R_0 {(1-r/R)h} \, dr = \alpha h(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = \alpha h(R - \frac{R^2}{2R}) = \alpha h\frac{R}{2} \\E2\pi rh = \frac{\alpha Rh}{2\epsilon_0}\\E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3. At the boundary where r = R:

$E(r=R) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R}) = \frac{\alpha}{4\pi \epsilon_0}\\E(r=R) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r} = \frac{\alpha}{4\pi \epsilon_0}$

As can be seen from above, two E-field values are equal as predicted.

4 0

3 years ago

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