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3 years ago

How many meters across is a road sign that has an angular size of 120 arcseconds and is 1 km away?

Physics

1 answer:

myrzilka [38]3 years ago

8 0

That’s the one I’m struggling withhhh aswellll

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63 J of heat are added to a closed system. The initial energy of the system is 58J, and the final internal energy is 93J. How mu

Nesterboy [21]

ΔU = Q + W

heat in, Q +
heat out, Q -
does work , W -
work in, W +

93-58 = 63 + W

35 = 63 + W

W = - 28 J (does work/being done by system)

7 0

2 years ago

;p;p;p;p;p;p;p;p;lol

Pachacha [2.7K]

Answer:

lol, I like math. and this question is awesome. lol.

Explanation:

So I like math because I am good at it. I hate ELA bc im bad at it.

7 0

3 years ago

Part A What is the resistance of a 4.4 m length of copper wire 1.3 mm n diameter? The resistivity of copper is 1.68x 10-8 Ω-m Ex

Natalija [7]

Explanation:

It is given that,

Length of the copper wire, l = 4.4 m

Diameter of copper wire, d = 1.3 mm = 0.0013 m

Radius of copper wire, r = 0.00065 m

The resistivity of the copper wire, $\rho=1.68\times 10^{-8}\ \Omega-m$

We need to find the resistance of the copper wire. It is given by :

$R=\rho\dfrac{l}{A}$

$R=1.68\times 10^{-8}\ \times \dfrac{4.4\ m}{\pi (0.00065)^2}$

R =0.055 ohms

So, the resistance of the copper wire is 0.055 ohms. Hence, this is the required solution.

7 0

3 years ago

A stone is dropped from rest from the top of a cliff into a pond below. If its initial height is 10 m, what is its speed when it

Brut [27]

Answer:

14 m/s

Explanation:

The motion of the stone is a free fall motion, so an accelerated motion with constant acceleration g = 9.8 m/s^2 towards the ground. So, we can use the following SUVAT equation:

$v^2 -u^2 = 2gh$

where

v is the final speed of the stone as it reaches the water

u = 0 is the initial speed

g = 9.8 m/s^2 is the acceleration

h = 10 m is the distance covered by the stone

Solving for v, we find

$v=\sqrt{u^2+2gh}=\sqrt{0+2(9.8 m/s^2)(10 m)}=14 m/s$

8 0

3 years ago

If the lily pads are spaced 2.4 m apart and Ferdinand jumps with a speed of 5 m/s taking 0.6 s to go from lily pad to lily pad,

Zina [86]

Answer:

$36.87^{\circ}$

Explanation:

v = Velocity of Ferdinand = 5 m/s

$\theta$ = Angle of jump

T = Time taken = 0.6 s

R = Distance between lily pads = 2.4 m

Horizontal range is given by

$R=v_xT\\\Rightarrow R=vcos\theta T\\\Rightarrow cos\theta=\dfrac{R}{vT}\\\Rightarrow \theta=cos^{-1}\dfrac{R}{vT}\\\Rightarrow \theta=cos^{-1}\dfrac{2.4}{5\times 0.6}\\\Rightarrow \theta=36.87^{\circ}$

The angle at which Ferdinand make each of his jumps is $36.87^{\circ}$

8 0

3 years ago