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3 years ago

PLEASE HELP. SO HARD. SCIENCE ASSIGNMENT. ASAP

Physics

1 answer:

Alja [10]3 years ago

7 0

World shall know pain im not helping figure it out urself + RATIO

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If the car passes point A with a speed of 20 m/s and begins to increase its speed at a constant rate of at = 0.5 m/s2 , determin

IceJOKER [234]

Answer:

$1.68 \frac{m}{s^2}$

Explanation:

Please find the image for the question as attached file.

Solution -

Given -

First of all we will calculate the velocity at point C,

As per newton's third law of motion-

$V_C^2 = V_A^2 + 2 a_t (S_C - S_A)\\$

Substituting the given values in above equation, we get -

$V_C^2 = 20^2 + 2*0.5*(100-0)\\V_C = 22.361 \frac{m}{s}$

Now we will determine the radius of curvature for the curve shown in the attached image

$Y = 16 - \frac{1}{625} X^2\\$

Differentiating on both the sides, we get -

$\frac{dy}{dx} = -3.2 (10^-3) X\\\frac{d^2y}{d^2x} = -3.2 (10^-3)\\Curve = \frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}} }{\frac{d^2y}{d^2x}} \\Curve = 312.5$ meter

Acceleration on curved path

$a = \frac{V_C^2}{Curve} \\a = \frac{22.361^2}{312.5} \\a= 1.60 \frac{m}{s^2}$

Final acceleration

$a_f = \sqrt{0.5^2 + 1.6^2} \\a_f = 1.68\frac{m}{s^2}$

5 0

3 years ago

The planet's hawks and block their near each other in the door again system the dworkin's have very advanced technology and a do

Virty [35]

Answer: A,B, and E

Explanation: Just checked I got them right:)

4 0

3 years ago

An object is projected upward from the surface of the earth with an initial speed of 3.7 km/s. find the maximum height it reache

Mariulka [41]

The kinetic energy of the object just after it starts its motion from the Earth surface is
$K= \frac{1}{2}mv^2$
where m is the object mass and $v=3.7 km/s=3700 m/s$ its initial speed.
When it reaches its maximum height, the object speed is zero and all its kinetic energy converted into gravitational potential energy, which is
$U=mgh$
where $g=9.81 m/s^2$ and h is the maximum height reached by the object.

Since the energy of the object must be conserved, K=U, therefore we can write
$\frac{1}{2}mv^2 =mgh$
and we can solve to find h, the maximum height:
$h= \frac{v^2}{2g}= \frac{(3700 m/s)^2}{2\cdot 9.81 m/s^2}= 6.98 \cdot 10^5 m=698 km$

5 0

4 years ago

As part of a physics experiment, you carry a bathroom scale calibrated in newtons onto an elevator and stand on it. At rest, you

mafiozo [28]

Answer: 708 N

Explanation:

Given

At rest, Elevator reads 588 N

When it starts moving upward at $2\ m/s^2$ , apparent weight changes

i.e. weight can be given by

$\Rightarrow W'=m(g+a)\\\Rightarrow W'=mg+mg\cdot \dfrac{a}{g}\\\\\Rightarrow W'=W\left(1+\dfrac{a}{g}\right)\\\\\Rightarrow W'=588\left(1+\dfrac{2}{9.8}\right)\\\\\Rightarrow W'=707.99\approx 708\ N$

The apparent weight is 708 N

5 0

3 years ago

Two identical guitar strings are prepared such that they have the same length ( 0.62 m ) and are under the same amount of tensio

Tanya [424]

Answer:

The speed is 401.76m/s

Explanation:

To find the fundamental frequency we use

3rd.harmonic - 2nd.harmonic = 324Hz

And we know that

f=v/2L

So v=f2L

v= 324 x 2 x 0.62= 401.76m/s

4 0

3 years ago