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Kryger [21]
3 years ago
6

Look at the diagram. Emily needs to complete the circuit. What two points should be connected to complete the circuit and make t

he bulb glow?

Physics
2 answers:
Tatiana [17]3 years ago
6 0

Answer:

C) Points 2 and 4

Explanation:

In a complete electric circuit, wires carry the current from the battery to the bulb and from the bulb to the battery. The wires should be attached to points 2 and 4 to make the bulb glow.

pogonyaev3 years ago
4 0

Answer:

The points 2 and 4 should be connected.

Explanation:

To complete the circuit, we need to connect the two points which when connected, encompass the battery and the bulb in the circuit. The points 2 and 4 do the job, since they connect the terminal of the battery and the terminal of the bulb, and thus complete the circuit.

Therefore,  the choice C is correct.

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A metal coin has a volume of 835 mm3 and a mass of 5.67 g. What is the density of the coin?
Lilit [14]
Density formula: d = M/V
So I think the answer would be 5.67/ 835 I am not sure of the answer I got confused (´∀`) but I hope it will help

3 0
3 years ago
Read 2 more answers
Light of wavelength 630 nm is incident on a long, narrow slit. Determine the angular deflection of the first diffraction minimum
asambeis [7]

Answer:

a) 1.8°

b) 0.18°

c) 0.018°

Explanation:

Wavelength (λ) = 630nm = 630 *10^-9m

The equation that describes the angular deflection of a dark band is

Wsin(βm) = mλ

w = width of the single slit

λ = wavelength of the light

βm = angular deflection of the mth dark band.

a) In order to get the angular deflection of the first dark band for a slit with 0.02mm width, substitute w = 0.02mm = 0.02*10^-3 , m = 1 , λ= 630*10^-9

0.02*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.02*10^-3

Sin(β1) = 0.0315

β1 = Sin^-1(0.0315)

= 1.8°

b) substitute w = 0.2mm = 0.2*10^-3 , m = 1 , λ= 630*10^-9

0.2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.2*10^-3

Sin(β1) = 0.00315

β1 = Sin^-1(0.00315)

= 0.18°

c) substitute w = 2mm = 2*10^-3 , m = 1 , λ= 630*10^-9

2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 2*10^-3

Sin(β1) = 3.15*10^-4

β1 = Sin^-1(3.15*10^-4)

= 0.018°

6 0
3 years ago
A plastic cup weighing 100 g floats on water so that 1/4 of the volume of the cup is immersed in water. How much volume of oil c
k0ka [10]

Answer:

Any floating object displaces a volume of water equal in weight to the object's MASS. ... If you place water and an ice cube in a cup so that the cup is entirely full to the ... If you take a one pound bottle of water and freeze it, it will still weigh one ... Fresh, liquid water has a density of 1 gram per cubic centimeter (1g = 1cm^3, ...

5 0
3 years ago
A ball is thrown upward at a 45° angle. Inthe absence of air resistance, the ballfollows aA. tangential curve.B. sine curve.C. p
Evgesh-ka [11]

As ball is projected up in air at an angle of 45 degree without any air resistance

Let the initial speed will be v

now we will have

In x direction

v_x = v cos45

in y direction

v_y = vsin45

now displacement in x direction

x = (vcos45)t + 0

displacement in y direction

y = (vsin45)t - \frac{1}{2}gt^2

now from above two equations we have

y = (vsin45)\frac{x}{vcos45} - \frac{1}{2}g(\frac{x}{vcos45})^2

y = xtan45 - \frac{1}{2v^2cos^245}gx^2

so above equation is a quadratic equation and hence it will be a parabolic curve

so correct answer will be

<em>C. parabolic curve.</em>

8 0
3 years ago
Narysuj wykres zależności v(t) jeśli w chwili początkowej t=0 V=10m/s w każdej sekundzie szybkość zmniejsza się o 1m/s . Po jaki
irina1246 [14]

1) See graph in attachment

2) 10 s

3) 50 m

Explanation:

1)

In this problem, we have an object initially moving with a velocity of

v = 10 m/s

when the time is

t = 0 s

Then, we are told that the speed of the object is decreasing by 1 m/s every  second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.

The result can be found in the graph in attachment.

Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to

a=1 m/s^2

which is equivalent to the gradient of the line in the velocity-time graph.

2)

In this part, we want to find after what time the body will stop its motion.

To do that, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

In this problem:

u = 10 m/s is the initial velocity of the body

a=-1 m/s^2 is the acceleration

v = 0 m/s, because we want to find the time T at which the body will stop

Re-arranging the equation, we find:

T=-\frac{u}{a}=-\frac{10}{-1}=10 s

3)

In order to find the total distance covered by the body during its accelerated motion, we have to use another suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem:

u = 10 m/s is the initial velocity

a=-1 m/s^2 is the acceleration

t = 10 s is the time it takes for the body to stop (found in part 2)

Solving for s, we find the distance covered:

s=(10)(10)+\frac{1}{2}(-1)(10)^2=50 m

7 0
3 years ago
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