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3 years ago

Select the statements that describe a vector. Check all that apply.

Physics

2 answers:

victus00 [196]3 years ago

5 0

Answer:

2,3,5

Explanation:

balu736 [363]3 years ago

3 0

Answer:

Vectors have a size and a direction

vectors can have positive or negative values

Explanation:

You might be interested in

10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????

KATRIN_1 [288]

Answer: Tension = 47.8N, Δx = 11.5× $10^{-6}$ m.

Tension = 95.6N, Δx = 15.4× $10^{-5}$ m

Explanation: A speed of wave on a string under a tension force can be calculated as:

$|v| = \sqrt{\frac{F_{T}}{\mu} }$

$F_{T}$ is tension force (N)

μ is linear density (kg/m)

Determining velocity:

$|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{0.00874 }$

$|v| =$ 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

$\Delta x = |v|.t$

$\Delta x = 9.35.10^{-2}*1.23.10^{-3}$

Δx = 11.5× $10^{-6}$

With tension of 47.8N, a pulse will travel Δx = 11.5× $10^{-6}$ m.

Doubling Tension:

$|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{2.0.00874 }$

$|v| = \sqrt{0.01568}$

|v| = 0.1252 m/s

Displacement for same time:

$\Delta x = |v|.t$

$\Delta x = 12.52.10^{-2}*1.23.10^{-3}$

$\Delta x =$ 15.4× $10^{-5}$

With doubled tension, it travels $\Delta x =$ 15.4× $10^{-5}$ m

4 0

4 years ago

A satellite m-500 kg orbits the earth at a distance d 215 km, above the surface of the planet. The radius of the earth is re 6.3

AnnZ [28]

Answer:

7.78 * 10³ m/s

Explanation:

Orbital velocity is given as:

v = √(GM/R)

G = 6.67 * 10^(-11) Nm/kg²

M = 5.98 * 10^(24) kg

R = radius of earth + distance of the satellite from the surface of the earth

R = 2.15 * 10^(5) + 6.38 * 10^(6)

R = 6.595 * 10^(6) m

v = √([6.67 * 10^(-11) * 5.98 * 10^(24)] / 6.595 * 10^(6))

v = √(6.048 * 10^7)

v = 7.78 * 10³ m/s

4 0

3 years ago

Calculate the resistance of an electric bulb which allows a 10 A current when connected to a 220 V power source

kodGreya [7K]

Resistance can be calculated by Ohm's law

As per ohm's law we will have

$V = i * R$

here we will have

voltage = 220 volts

current = 10 A

So by the above formula we will have

$220 = 10* R$

$R = 22 ohm$

So resistance of the bulb is 22 ohm.

4 0

4 years ago

A 22.0 ohm and 75.0 ohm resistor are in parallel, connected to a 5.00 v battery. how much current flows out of the battery

Julli [10]

Answer:

294 mAmps

Explanation:

The two resistors are equal to 22 * 75 / ( 22+ 75) = 17.01 ohm

V = IR

V/R = I

5 / 17.01 = .294 A

5 0

2 years ago

An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens.

CaHeK987 [17]

A) 11.1 cm

We can find the focal length of the lens by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p = 16.0 cm is the distance of the object from the lens

q = 36.0 cm is the distance of the image from the lens (taken with positive sign since it is on the opposide side to the image, so it is a real image)

Solving the equation for f:

$\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{36.0 cm}=0.09 cm^{-1}\\f=\frac{1}{0.09 cm^{-1}}=11.1 cm$

B) Converging

The focal length is:

- Positive for a converging lens

- Negative for a diverging lens

In this case, the focal length is positive, so it is a converging lens.

C) 18.0 mm

The magnification equation states that:

$\frac{h_i}{h_o}=-\frac{q}{p}$

where

$h_i$ is the heigth of the image

$h_o$ is the height of the object

$q=36.0 cm$

$p=16.0 cm$

Solving the formula for $h_i$ , we find

$h_i = -h_o \frac{q}{p}=-(8.00 mm)\frac{36.0 cm}{16.0 cm}=-18.0 mm$

So the image is 18 mm high.

D) Inverted

From the magnification equation we have that:

- When the sign of $h_i$ is positive, the image is erect

- When the sign of $h_i$ is negative, the image is inverted

In this case, $h_i$ is negative, so the image is inverted.

4 0

4 years ago