Please help me answer this question

1. What is the momentum of a 0.15 kg arrow that is traveling at 120 m/s?

Korvikt [17]

For this case we have that by definition, the momentum equation is given by:

$p = m * v$

Where:

m: It is the mass

v: It is the velocity

According to the data we have:

$m = 0.15 \ kg\\v = 120 \frac {m} {s}$

Substituting:

$p = 0.15 * 120\\p = 18 \frac {kg * m} {s}$

On the other hand, if we clear the variable "mass" we have:

$m = \frac {p} {v}$

According to the data we have:

$p = 250 \frac {kg * m} {s}\\v = 5 \frac {m} {s}\\m = \frac {250} {5}\\m = 50 \ kg$

Thus, the mass is $50 \ kg$

Answer:

$p = 18 \frac {kg * m} {s}\\m = 50 \ kg$

3 0

3 years ago

A force in the x direction acts on a particle moving also in the x direction, producing a potential energy U(x)=Ax4 where A=0.63

poizon [28]

Answer:

The magnitude of the force is 1.29*10^-3N in the positive x direction

Explanation:

In order to calculate the magnitude and direction of the force, you take into account that the force is the space derivative of the potential enrgy, as follow:

$F(x)=-\frac{dU(x)}{dx}$ (1)

where:

$U(x)=Ax^4\\\\A=0.0630\frac{J}{m^4}$

You replace the expression for U into the equation (1) and solve for F:

$F(x)=-\frac{d}{dx}[Ax^4]=-4Ax^3$ (2)

The force on the particle, for x = -0.080m is:

$F=-4(0.630\frac{J}{m^4})(-0.0800m)^3=1.29*10^{-3}N$

The magnitude of the force is 1.29*10^-3N in the positive x direction

7 0

3 years ago

Which method will correctly determine whether the forces on an object are balanced or unbalanced?

Morgarella [4.7K]

<span>Subtract the forces in the horizontal direction from the forces in the vertical direction.</span>

4 0

3 years ago

Read 2 more answers

The magnitude J of the current density in a certain wire with a circular cross section of radius R = 2.11 mm is given by J = (3.

oksano4ka [1.4K]

Answer:

$i = 2.84 \times 10^{-3} A$

Explanation:

As we know that current density is ratio of current and area of the crossection

now we have

$J = \frac{di}{dA}$

so the current through the wire is given as

$i = \int J dA$

now we have

$i = \int_{0.921R}^R J dA$

here we have

$J = (3.25 \times 10^8)r^2$

now plug in the values in above equation

$i = \int_{0.921R}^R (3.25 \times 10^8)r^2 2\pi r dr$

now we have

$i = \int_{0.921R}^R 2\pi (3.25 \times 10^8)r^3 dr$

$i = (2.04 \times 10^9) \frac{r^4}{4}$

now plug in both limits as mentioned

$i = (2.04 \times 10^9)(\frac{R^4}{4} - \frac{(0.921R)^4}{4})$

$i = (2.04\times 10^9)(0.07 R^4)$

here R = 2.11 mm

$i = (2.04 \times 10^9)(0.07 (2.11 \times 10^{-3})^4)$

$i = 2.84 \times 10^{-3} A$

8 0

3 years ago

A 6 ft tall person walks away from a 10 ft lamppost at a constant rate of 5 ft/s. What is the rate (in ft/s) that the tip of the

nalin [4]

Answer:

12.5 ft/s

Explanation:

Height of person = 6 ft

height of lamp post = 10 ft

According to the question,

dx / dt = 5 ft/s

Let the rate of tip of the shadow moves away is dy/dt.

According to the diagram

10 / y = 6 / (y - x)

10 y - 10 x = 6 y

y = 2.5 x

Differentiate both sides with respect to t.

dy / dt = 2.5 dx / dt

dy / dt = 2.5 (5) = 12.5 ft /s

8 0

4 years ago