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3 years ago

11

Newton’s first law of motion states that an object's motion remains the same unless a force acts upon it. What would be an examp

le of this law in action?

1 answer:

LekaFEV [45]3 years ago

5 0

Answer:

A driver breaks suddenly and the passengers are flung forward

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A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. T

slamgirl [31]

Answer:

$u=14.48m/s$

Explanation:

From the question we are told that:

Height of window $h=2m$

Height of window off the ground $h_g=7.5m$

Time to fall and drop $t=1.3s$

Generally the Newton's equation motion is mathematically given by

$s=ut+\frac{1}{2}at^2$

Where

$h=ut+\frac{1}{2}at^2$

$2=u1.3-\frac{1}{2}*9.8*1.3^2$

$2=u1.3-8.281$

$u=7.91m/s^2$

Generally the Newton's equation motion is mathematically given by

$2as=v^2-u^2$

Where

$-2gh_g=v^2-u^2$

$-2*9.8*7.5=(7.91)^2-u^2$

$-147=62.5681-u^2$

$u=\sqrt{209.5681}$

$u=14.48m/s$

Therefore the ball’s initial speed

$u=14.48m/s$

8 0

3 years ago

You move to Mars and as a momento from Earth take your Ma-maw's mercury barometer. You place it outside in the Martian atmospher

oksano4ka [1.4K]

Answer:

Explanation:

No, the reading is not expected to be accurate. This is because Relative to Earth, the air on Mars is extremely thin. The Martian atmosphere is primarily carbon dioxide with a much lower surface pressure, and Mars does not have oceans and an Earthlike hydrological cycle so latent heat release is not as important as it is for Earth.

8 0

3 years ago

The main illustration in the video shows the life track of a one-solar mass star. Each point along this track represents _______

REY [17]

Each point along the track of one solar mass star represents the star's surface temperature and luminosity at one time.

<h3>What is the one-solar mass star?</h3>

A star having a mass equal to the mass of the Sun is called a one-solar mass star.

Its life track shows the luminous intensity as well as the surface temperature.

Learn more about one-solar mass star.

brainly.com/question/14984575

#SPJ1

7 0

2 years ago

5. Find the volume of the composite space figure to the nearest whole number.

Dmitry [639]

The shape is missing but let's consider it a semi-cylinder attached to the rectangular prism.
Given:
radius = 4.5 mm
<span>Height = 11 mm </span>

<span>Volume of cylinder = (1/2)(pi)(4.5)^2(11) (the shape is divided into half)
V = 349.89 mm cubed
Volume of prism = L x W x H
= 9 x 11 x 6
= 594 mm cubed
Total volume of the composite shape = 111.375 + 594
= 943.89 mm cubed
Rounded answer = 944 mm cubed.</span>

6 0

4 years ago

Determine the kinetic energy of the ball immediately after it is hit. (You must provide an answer before moving to the next part

earnstyle [38]

The question is incomplete. Here is the complete question.

A baseball palyer hits a 5.1 oz baseball with an initial velocity of 140ft/sat an angle of 40° with the horizontal as shown. Determine

a) The kinetic energy of the ball immediately after it is hit

b) The kinetic energy of the ball when it reaches its maximum height

c) The maximum height above the ground reached by the ball.

Answer: a) KE = 131.64 J

b) KE = 0

c) h = 126 ft

Explanation: <u>Kinetic</u> <u>energy</u> is the energy an object posses due to its motion. It can be calculated as $KE=\frac{1}{2}mv^{2}$

a) Kinetic energy's unit is Joule. So, we have to transform ounce in kg and ft/s in m/s for the units to correspond:

m = 5.1(0.02835)

m = 0.1445 kg

v = 140 ft = 42.67 m/s

Then, kinetic energy is

$KE=\frac{1}{2}(0.1445)(42.67)^{2}$

KE = 131.64 J

Kinetic energy immediately after the ball is hit is 131.64 J.

b) At its maximum height, the ball has its highest potential energy. Because of the law of conservation of energy, when potential energy is maximum, kinetic energy is minimum and vice-versa. So, at the maximum height, kinetic energy is 0.

c) This type of motion is <u>projectile</u> <u>motion</u>. The maximum height on a projectile motion can be determined by

$v_{y}^{2}=v_{0y}^{2}-2g\Delta y$

When h is maximum, $v_{y}=0$

Velocity of the ball has an angle with the horizontal, so initial velocity at the y-axis is

$v_{0y}=v_{0}sin(\theta)$

Substituting and solving

$v_{y}^{2}=v_{0}^{2}sin^{2}(\theta)-2gh$

$0=(42.67)^{2}sin^{2}(40)-2(9.8)h$

$19.6h=(42.67)^{2}(0.643)^{2}$

$h=\frac{(1820.73)(0.4132)}{19.6}$

h = 38.4 m

Transforming into ft: h = 126 ft

The maximum height above the ground reached by hte ball is 126 feet.

6 0

3 years ago