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3 years ago

Hurricane Katrina and hurricane Rita are similar in what way? (I will give brainliest to the correct answer) :)

Physics

2 answers:

scoundrel [369]3 years ago

5 0

Answer:

both same way

Explanation:

RoseWind [281]3 years ago

4 0

Answer:

Rita and Katrina both followed similar paths into the Gulf.

Explanation:

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A toy car moves 8 min 4 s at the constant velocity. What is the car's velocity?

Schach [20]

Explanation:

Formula:

$velocity = (d \div t)$

d = distance given

t = the amount of time given

Substitute the given values into the formula for velocity:

$v = 8 \div 4$

velocity is shortened for v.

8 (distance) divided by 4 (time) equals the velocity.

Solve:

$2 = 8 \div 4$

The velocity of the toy car equals: B. 2 m/s.

6 0

3 years ago

If you had started with a larger mass, how would the half-life change?

iogann1982 [59]

Answer:

There is no change, unless your mass is somehow at the quantum level, at which the concept of half-life breaks down.

Half life is a property of the specific radioactive isotope...NOT of the initial sample's mass.

3 0

3 years ago

Can someone help me please

stealth61 [152]

B hope this helps you!!!

4 0

3 years ago

You are driving your car along a country road at a speed of 27.0 m/s. as you come over the crest of a hill, you notice a farm tr

Bezzdna [24]

speed of the car = 27 m/s

speed of truck ahead = 10 m/s

relative speed of car with respect to truck

$v_r = 27 - 10 = 17 m/s$

relative deceleration of car

$a_r = -7 m/s^2$

now the distance before they stop with respect to each other is given by

$v_f^2 - v_i^2 = 2 a d$

$0 - 17^2 = 2 *(-7)*d$

$d = 20.6 m$

so it will come at the same speed of truck after 20.6 m distance and hence it will not hit the truck as the distance of the truck is 25 m from car

Part b)

Distance traveled by car before it stops is given by

$v_f^2 - v_i^2 = 2 a s$

$0^2 - 27^2 = 2 * (-7)* s$

$s = 52.1 m$

so it will stop after it will cover total 52.1 m distance

Part c)

time taken by the car to stop

$v_f - v_i = at$

$0 - 27 = (-7) * t$

$t = 3.86 s$

now the distance covered by truck in same time

$d = 3.86 * 10 = 38.6 m$

now after the car will stop its distance from the truck is

$D = 25 + 38.6 - 52.1 = 11.5 m$

so the distance between them is 11.5 m

6 0

3 years ago

An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explode

sasho [114]

The x- and y-coordinates are 9142.57 m and -304.425 m

Explanation:

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell $v_{0}$ in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

$v_{x}=v_{0} \times \cos \theta$

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of $v_{x}$ and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

$v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}$

$v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}$

For motion with constant acceleration, we know

$s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}$

Along the horizontal, x-axis, we might write this as

$x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}$

Measuring distances relative to the firing point means

$x_{0}=0$

we know that,

$a_{x}=0$

or,

$v_{x}=v_{x 0}=\text { constant }$

By applying the values, we get,

$x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}$

The acceleration of gravity is vertically downward and is $g=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

$y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}$

we know, $y_{0}=0$ and $a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , so,

$y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)$

y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m

7 0

3 years ago