All categories

3 years ago

Plzz answer this questions will mark as brainlist

Physics

1 answer:

EastWind [94]3 years ago

4 0

Answer:

1. 19.28 secs

2. 154.22 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 16 m/s

Final velocity (v) = 0

Force (F) = 1000 N

Mass (m) = 1200 Kg

Time (t) =..?

Distance (s) =...?

Next, we shall determine the acceleration of the car. This can be obtained as follow:

Force (F) = 1000 N

Mass (m) = 1200 Kg

Acceleration (a) =.?

Force (F) = mass (m) x acceleration (a)

F = ma

1000 = 1200 x a

Divide both side by 1200

a = 1000/1200

a = 0.83 m/s²

Since the car is coming to rest, it means it is decelerating. Therefore, the acceleration is – 0.83 m/s²

1. Determination of time taken for the car to halt i.e stop. This can be obtained as follow:

Initial velocity (u) = 16 m/s

Final velocity (v) = 0

acceleration (a) = – 0.83 m/s²

Time (t) =.?

v = u + at

0 = 16 + (–0.83 x t)

0 = 16 – 0.83t

Rearrange

0.83t = 16

Divide both side by 0.83

t = 16/0.83

t = 19.28 secs.

Therefore, the time taken for the car to halt is 19.28 secs.

2. Determination of the distance travelled by the car before coming to rest. This can be obtained as follow:

Initial velocity (u) = 16 m/s

Final velocity (v) = 0

acceleration (a) = – 0.83 m/s²

Distance (s) =..?

v² = u² + 2as

0 = 16² + (2 x –0.83 x s)

0 = 256 – 1.66s

Rearrange

1.66s = 256

Divide both side by 1.66

s = 256/1.66

s = 154.22 m

Therefore, the distance travelled by the car before coming to rest is 154.22 m.

You might be interested in

A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur

Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0

3 years ago

The product of voltage times amperage is known as what? amperes current resistance power

UNO [17]

The product of (voltage) times (current, in Amperes) is POWER.

5 0

3 years ago

Read 2 more answers

What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

Wittaler [7]

Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

$\cos \theta = \frac{adjacent}{hypotenuse}$

$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

$r_{71.9^{\circ}}$ can be isolated from equation 3:

$r_{71.9^{\circ}} = r_{e} \cos \theta$ (4)

$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

$v = \frac{2 \pi (1.97x10^{6} m)}{24h}$

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

$T = 24h . \frac{3600s}{1h}$ ⇒ $84600s$

$v = \frac{2 \pi (1.97x10^{6} m)}{84600s}$

$v = 146.31m/s$

Finally, equation 1 can be used:

$a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}$

$a = 0.010m/s^{2}$

Hence, the centripetal acceleration is $0.010m/s^{2}$

To given the centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ it is gotten:

$\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}$

6 0

3 years ago

An object is placed at a far distance in front of a thin converging lens. Then the object (1 point) moves gradually toward the l

xeze [42]

Answer:

real, and then virtual

Explanation:

A converging lens is known as convex lens. This lens is called converging lens because it converges all light rays incident on the lens and parallel to the principal axis at the focus.

The nature of image formed by objects placed in front of this lens as mostly REAL IMAGES. The image formed becomes virtual only when the object is almost in close contact with the lens.

Based on the explanation, it can be deduced that an object placed far from a convex lens forms real images but as we move closer to the lens (almost touching the lens), the image formed overtime tends to be virtual.

8 0

3 years ago

A constant friction force of 25 N acts on a 65-kg skier for 15 s on level snow. What is the skier’s change in velocity

nikdorinn [45]

Answer:

$\Delta v=5.77m/s$

Explanation:

Newton's 2nd Law relates the net force F on an object of mass m with the acceleration a it experiments by F=ma. In our case the net force is the friction force, since it's the only one the skier is experimenting horizontally and the vertical ones cancel out since he's not moving in that direction. Our acceleration then will be:

$a=\frac{F}{m}$