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Westkost [7]
3 years ago
9

SOHCAHTOA MATH REVIEW Find the angle x from the diagram above

Physics
1 answer:
mestny [16]3 years ago
5 0

Explanation:

Hey there!

Here,

It is a Right angled triangle.

Now, Taking angle x as a reference angle.

We get,

perpendicular (p) = 24.

Hypotenuse (h) = 29.

Using sin ratio we get.

\sin( \alpha )  =  \frac{p }{h}

Now, keep values and simplify them.

\sin(x)  =  \frac{24}{29}

sinx = 0.8275.

x =  { \sin }^{  - 1}  (0.8275)

Therefore the value of x is 55.842°. Or 55°.

<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

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vredina [299]

Answer:

0.5 m/s2

Explanation:

accelration formula : final velocty - starting velocity divided by time

5 0
2 years ago
A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the
lana [24]
(a) Let's convert the final speed of the car in m/s:
v_f = 61 km/h = 16.9 m/s
The kinetic energy of the car at t=19 s is
K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
P= \frac{W}{\Delta t}
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W

(c) The instantaneous power is given by
P_i = Fv_f
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
a= \frac{v_f-v_i}{\Delta t}=  \frac{16.9 m/s}{19 s}=0.89 m/s^2
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W
5 0
3 years ago
Mechanical energy is a term that is used to describe
larisa86 [58]
The sum of potential energy<span> and kinetic </span><span>energy.
Hope I helped!</span>
7 0
3 years ago
An electron is released from rest in a weak electric field given by E =-2.30 x 10-10 N/Cj. After the electron has traveled a ver
Goshia [24]

Explanation:

It is given that,

An electron is released from rest in a weak electric field of, E=2.3\times 10^{-10}\ N/C

Vertical distance covered, s=1\ \mu m=10^{-6}\ m

We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

v^2-u^2=2as

v^2=2as.............(1)

Electric force is F_e and force of gravity is F_g. As both forces are acting in downward direction. So, total force is:

F=mg+qE

F=9.1\times 10^{-31}\times 9.8+1.6\times 10^{-19}\times 2.3\times 10^{-10}

F=4.57\times 10^{-29}\ N

Acceleration of the electron, a=\dfrac{F}{m}

a=\dfrac{4.57\times 10^{-29}\ N}{9.1\times 10^{-31}\ kg}

a=50.21\ m/s^2

Put the value of a in equation (1) as :

v=\sqrt{2as}

v=\sqrt{2\times 50.21\times 10^{-6}}

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So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.

6 0
3 years ago
A train moving near the speed of light enters a tunnel. According to a person standing in the middle of the tunnel, the back end
Neporo4naja [7]

Answer: The person sitting in the middle of the train sees the back of the train enter ing the tunnel before the front end comes out.

Explanation:

4 0
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