A car moves at a constant speed of 10 m/s. If the car doesn't accelerate during the next 40 s how far will it go?

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

3 years ago

Marvin Martian is standing on planet Potatoine.

prisoha [69]

Answer:

W = 113.98 N

Explanation:

Given that,

Radius of Potatoine $R=6.4\times 10^6\ m$

Mass of Potatoine, $M=2\times 10^{24}\ kg$

Mass of Marvin, m = 35 kg

We need to find his weight on Potatoine. Weight of an object is given by :

W = mg

g is acceleration due to gravity, $g=\dfrac{GM}{R^2}$

So,

$W=\dfrac{GMm}{R^2}\\\\W=\dfrac{6.67\times 10^{-11}\times 35\times 2\times 10^{24}}{(6.4\times 10^6)^2}\\\\W=113.98\ N$

So, his weight on Potatoine is 113.98 N.

4 0

2 years ago

A swimmer moves through the water at 2.5 meters per second and experiences a drag force of 240 N. How much power is she generati

mote1985 [20]

-- There's a force of 240N pushing her backwards.

-- She's maintaining a steady speed (of 2.5 m/s) .

-- In order to maintain a steady speed (no acceleration),
the forces on her must be balanced. So she's maintaining
a steady force of 240N forward.

-- Every time she moves 1 m forward, she does work of
(force) x (distance) = 240 joules.

-- She moves 2.5 meters forward every second.
So she's doing (240 x 2.5) = 600 joules of work every second.

-- 600 joules per second = 600 watts .

6 0

3 years ago

Human hox genes are in four clusters: hoxa, hoxb, hoxc, and hoxd, for a total of 39 genes. besides conserving the clustering of

Jet001 [13]

Human hox genes are in four clusters: HOXA, HOXB, HOXC, and HOXD for a total of 39 genes. besides conserving the clustering of genes, evolution has also conserved the order of genes within them.

What is a gene?

A gene is the basic physical and functional unit of heredity. Genes are made up of DNA. Some genes helps to make molecules called proteins. However, many genes do not synthesis for proteins. Every person has two copies of each gene, one inherited from each parent. Most genes are the same in all people, but a small number of genes (less than 1 percent of the total) are slightly different among the people. Alleles are forms of the same gene with small differences in their sequence of DNA bases. These small differences contribute to each person’s unique physical features.

To learn more about genes, refer to:

brainly.com/question/8832859

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5 0

8 months ago

Approximating the eye as a single thin lens 2. 75 cmcm from the retina, find the eye's near-point distance if the smallest focal

RSB [31]

The near-point distance of the eye is 11 cm.

<h3>What is the eye's near-point distance?</h3>

The near-point distance of the eye is the closest possible distance an object can be from the eye in order for its image to be formed on the retina. It can also be termed the closest distance of accommodation.

The near-point distance of the eye in the given scenario can be calculated using the lens formula given below:

1/f = 1/v + 1/u

where;

f = focal length

v = image distance

u = object distance

From the data provided;

f = 2.20 cm

v = 2.75 cm

u = ?

Solving for u:

1/u = 1/f - 1/v

1/u = 1/2.20 - 1/2.75

1/u = 0.91

u = 11 cm

In conclusion, the lens formula is used to determine the eye's near-point distance.

Learn more about eye's near-point distance at: brainly.com/question/16391605

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5 0

1 year ago