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SSSSS [86.1K]
3 years ago
14

What is the acceleration of a 50 kg object pushed with a net force of 500 newtons?

Physics
1 answer:
vodomira [7]3 years ago
3 0
Using the formula F=ma
500N=50kg (a)
a= 10 m/s^2
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In the same configuration of the previous problem 3, four long straight wires are perpendicular to the page, and their cross sec
faust18 [17]

Complete Question:

In the same configuration of the previous problem 3, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a = 13.5 cm. Each wire carries 7.50 A, and the currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3.

a) Draw a diagram in a (x,y) plane of the four wires with wire 4 perpendicular to the origin. Indicate the current's directions.

b) Draw a diagram of all magnetic fields produced at the position of wire 3 by the other three currents.

c) Draw a diagram of all magnetic forces produced at the position of wire 3 by the other three currents.

d) What are magnitude and direction of the net magnetic force per meter of wire length on wire 3?

Answer:

force, 1.318 ₓ 10⁻⁴

direction, 18.435°

Explanation:

The attached file gives a breakdown step by step solution to the questions

7 0
3 years ago
How many earths could fit inside jupiter
Snezhnost [94]

1,300 or more.

hope this helped :)

8 0
3 years ago
PLEASE HELP ASAP!!!
musickatia [10]

Answer:

22.2 W

Explanation:

First of all, we calculate the work done by moving the wagon, using the formula:

W=Fdcos \theta

where

F = 20 N is the magnitude of the force

d = 1000 m is the displacement of the wagon

\theta=0^{\circ} is the angle between the direction of the force and of the displacement (assuming the force is applied in the direction of motion)

Substituting, we find

W=(20)(1000)=20,000 J

Now we can find the power generated, which is equal to the ratio between the work done and the time taken:

P=\frac{W}{t}

where

W = 20,000 J

t = 15 min = 900 s

Substituting,

P=\frac{20000}{900}=22.2 W

And the same value in Joules/second (remember that 1 Watt = 1 Joule/second)

5 0
3 years ago
Objective lenses are contained in a ____________ that can be turned to put a particular objective lens in place to be used.
Masteriza [31]

Answer:

Revolving nosepiece

Explanation:

The revolving nosepiece is one of the parts of a microscope, used for holding the objective lenses. They can be turned to put a particular objective lens in place to be used in order to vary magnification.

6 0
3 years ago
A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c
ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

Integrate on both the sides

\int _{0}^{i}di =9\int _{0}^{t}dt

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

B = \frac{\mu_{0}i}{2\pi r}

B = \frac{\mu_{0}\times 9t}{2\pi r}

(b)

Magnetic flux,

\phi=\int B\times a dr

\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr

\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )

\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)

\phi = 1.89 \times 10^{-7}t

(c)

R = 3 ohm

e = -\frac{d\phi}{dt}

magnitude of voltage is

e = 1.89 x 10^-7 V

induced current, i = e / R = (1.89 x 10^-7) / 3

i = 6.3 x 10^-8 A

8 0
3 years ago
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