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Arlecino [84]
2 years ago
11

What is the potential difference across lamp 1

Physics
1 answer:
Alex73 [517]2 years ago
7 0

Answer:

1+1=2

Explanation:

You just need to add and voila the answer

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A tuning fork with a frequency of 404 Hertz is struck with a soft hammer. The fork is struck a second time by the
Bas_tet [7]

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:D :D :D :D

Explanation:

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3 years ago
Which statement describes an example of static electricity?
lana [24]

Answer: Negatively charged particles are repelled by other negatively charged particles

Explanation:

7 0
3 years ago
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What is the formula for instant velocity when I have time and distance?
Tomtit [17]

Answer:

v(t)= (d/dt)x(t)

Explanation:

The instantaneous velocity of an object is the limit of the average velocity as the elapsed time approaches zero, or the derivative of x with respect to t. Like average velocity, instantaneous velocity is a vector with dimension of length per time. The instantaneous velocity at a specific time point  t 0  is the rate of change of the position function, which is the slope of the position function  

x ( t ) at  t 0 .

3 0
3 years ago
A stone is thrown with an initial speed of 12 m/s at an angle of 30o above the horizontal from the top edge of a cliff. If it ta
kherson [118]

Answer:

d=58m

Explanation:

From the question we are told that:

Initial Speed U=12m/s

Time T=5.6s

Angle \theta=30

Generally the  Newton's equation for motion is mathematically given by

d=d'+ut+\frac{at^2}{2}

d=12cos30*5.6

d=58m

8 0
3 years ago
A stone is thrown upward from the top of a building at an angle of 30° to the horizontal and with an initial speed of 20 m/s. Th
Hitman42 [59]

Answer:

Explanation:

Given

inclination \theta =30^{\circ}

initial speed u=20\ m/s

Point of release is 45 m above the ground

Considering stone to be a projectile, so time taken by projectile for its zero vertical displacement is

t_1=\frac{2u\sin \theta }{g}

t_1=\frac{2\times 20\times \sin 30}{10}

t_1=2\ s

Now after completing zero vertical displacement , stone needs to travel another 45 m in downward direction with initial speed u=20\sin 30

h=u_yt+\frac{1}{2}a_yt^2

where, h=height

u_y=vertical velocity

a_y=vertical acceleration

t_0=time

45=20\sin 30+\frac{1}{2}(9.8)(t_0)^2

t_0^2=\frac{70}{9.8}

t_0=2.64\ s

thus total time time required is t=t_0+t_1=2.64+2=4.64\ s

vertical velocity just before hitting

v_y=\sqrt{u_y^2+2\times a_y\times s}

v_y=\sqrt{10^2+2\times 10\times 45}

v_y=\sqrt{1000}=31.622\ m/s

Horizontal velocity v_x=u\cos 30=17.32\ m/s

Net velocity Just before hitting =\sqrt{v_x^2+v_y^2}

=\sqrt{(17.32)^2+(31.62)^2}

=\sqrt{1299.82}=36.05\ m/s

                 

7 0
3 years ago
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