Ask question

Ask question

All categories

2 years ago

8

an 2-kg object is moving horizontally with a speed of 4m/s. how much net force os required to keep the object movong with the sa

me speed and in the same direction

1 answer:

Zepler [3.9K]2 years ago

7 0

In order to keep the object moving in the same direction and with same speed, its initial and final momentum should be zero.
∴ Net force on the object should be zero.

You might be interested in

A 1.47-newton baseball is dropped from a height of 10.0 meters and falls through the air to the ground. The kinetic energy of th

vagabundo [1.1K]

Answer:

The maximum amount of mechanical energy converted to internal energy during the fall is 26.7 joules

Explanation:

Potential Energy (PE) = weight of baseball × height = 1.47N × 10m = 14.7Nm = 14.7 joules

Kinetic Energy (KE) = 12 joules

Maximum amount of mechanical energy converted to internal energy during the fall = PE + KE = 14.7 joules + 12 joules = 26.7 joules

8 0

1 year ago

Read 2 more answers

A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a consta

-Dominant- [34]

The car would go from zero to 58.0 mph in 2.6 sec.

Since the force on the car is constant, therefore the acceleration of the car would also be constant.

Now for constant acceleration we can use the equation of motion

Using first equation of motion to calculate the acceleration of the car

v=u+at

29=0+a×1.30 ...... Eq. (1)

Again using the first equation of motion

58=0+a*t ....... Eq. (2)

Dividing eq. (2) with equation 1

t=2×1.3

t=2.6 sec

7 0

2 years ago

A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s of its motion, the vertical accel

Mazyrski [523]

$a = 2.7t$
$v = \int\limits^t_0 {2.7t} \, dt = \frac{2.7}{2} t^2$
$x = \int\limits^t_0 {\frac{2.7}{2} t^2} \, dt = \frac{2.7}{6} t^3$

Solve for v with x = 325.

5 0

2 years ago

Read 2 more answers

Calculate the mag-netic field (magnitude and direc-tion) at a point p due to a current i=12.0 a in the wire shown in fig. p28.68

creativ13 [48]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The magnitude is $B= 4.2 *10^ {-6}T$ , the direction is into the page

Explanation:

From the question we are told that

The current is $i = 12.0 A$

The radius of arc bc is $r_{bc} = 30.0 \ cm =\frac{30}{100} = 0.3m$

The radius of arc da is $r_{da} = 20.0 \ cm = \frac{20}{100} = 0.20 \ m$

The length of segment cd and ab is = $l = 10cm = \frac{10}{100} = 0.10 m$

The objective of the solution is to obtain the magnetic field

Generally magnetic due to the current flowing in the arc is mathematically represented as

$B = \frac{\mu_o I}{4 \pi r}$

Here I is the current

$\mu_o$ is the permeability of free space with a value of $4\pi *10^{-7}T \cdot m/A$

r is the distance

Considering Arc da

$B_{da} = \frac{\mu_o I}{4 \pi r_{da}} \theta$

Where $\theta$ is the angle the arc da makes with the center from the diagram its value is $\theta = 120^o = 120^o * \frac{\pi}{180} = \frac{2\pi}{3} rad$

Now substituting values into formula for magnetic field for da

$B_{da} = \frac{4\p *10^{-7} * 12}{4 \pi (0.20)}[\frac{2 \pi}{3} ]$

$= \frac{10^{-7} * 12}{0.20} * [\frac{2 \pi}{3} ]$

$B_{da}= 12.56*10^{-6} T$

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to da is into the pages of the paper

Considering Arc bc

$B_{bc} = \frac{\mu_o I}{4 \pi r_{bc}} \theta$

Substituting value

$B_{bc} = \frac{4 \pi *10^{-7} * 12}{4 \pi (0.30)} [\frac{2 \pi}{3} ]$

$B_{bc}= 8.37*10^{-6}T$

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to bc is out of the pages of the paper

Since the line joining P to segment bc and da makes angle = 0°

Then the net magnetic field would be

$B = B_{da} - B{bc}$

$= 12.56*10^{-6} - 8.37*10^{-6}$

$= 4.2 *10^ {-6}T$

Since $B_{da} > B_{bc}$ then the direction of the net charge would be into the page

3 0

2 years ago

The absolute pressure at a given point is 50 mmHg. Express this pressure in kPa, kPa gage and atm if atmospheric pressure is 80

miv72 [106K]

Answer:6.66 kPa

Explanation:

Given

Absolute pressure is 50 mm of Hg

atmospheric pressure=80 kPa

we know that

Absolute Pressure=Gauge pressure+vacuum Pressure

50 mm of Hg is equivalent to 6.66 kPa

Thus

Absolute Pressure is 6.66 kPa

or it can be said that it is 73.34 kPa vacuum pressure

4 0

2 years ago