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Gelneren [198K]
3 years ago
11

SOMEONE HELP ME!!!!!

Physics
2 answers:
aev [14]3 years ago
8 0

Answer:

The angle of reflection is 180 degrees because if you mapped it on a coordinate plane, both the real object and the image in the mirror will be on opposite sides.

katen-ka-za [31]3 years ago
7 0

Answer:

angle of refraction is the same as angle of incidence

Explanation:

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What happens to a circuit's resistance (R), voltage (V), and current (1) when
Naya [18.7K]

Answer:

B

Explanation:

Hope this helps

4 0
2 years ago
Read 2 more answers
A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr
KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s

Time the parachutist falls without friction is 3.19 seconds

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2

Now the initial velocity of the last half height will be the final velocity of the first half height.

v=u+at\\\Rightarrow v=at

Since the height are equal

\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0

t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2

Magnitude of acceleration is -43200 ft/s²

5 0
3 years ago
1. A limiting factor for using nuclear energy is the
Goryan [66]

Answer:

waste

Explanation:

took da test. nhgvgg

8 0
3 years ago
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The force of gravitation between two spherical bodies is Gm1 m2 /r2, where r is separation between their dash
AnnZ [28]

Explanation:

F = Gm1m2/r^2

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6 0
3 years ago
Read 2 more answers
A high-jump athlete leaves the ground, lifting her center of mass 1.8 m and crossing the bar with a horizontal velocity of 1.4 m
Romashka [77]

Answer:

The minimum speed when she leave the ground is 6.10 m/s.

Explanation:

Given that,

Horizontal velocity = 1.4 m/s

Height = 1.8 m

We need to calculate the minimum speed must she leave the ground

Using conservation of energy

K.E+P.E=P.E+K.E

\dfrac{1}{2}mv_{1}^2+0=mgh+\dfrac{1}{2}mv_{2}^2

\dfrac{v_{1}^2}{2}=gh+\dfrac{v_{2}^2}{2}

Put the value into the formula

\dfrac{v_{1}^2}{2}=9.8\times1.8+\dfrac{(1.4)^2}{2}

\dfrac{v_{1}^2}{2}=18.62

v_{1}=\sqrt{2\times18.62}

v_{1}=6.10\ m/s

Hence, The minimum speed when she leave the ground is 6.10 m/s.

6 0
3 years ago
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