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3 years ago

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PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE H

ELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP
I NEED TO GIVE THIS IN TODAY!!!

1 answer:

Masteriza [31]3 years ago

5 0

Answer:

a) greater

smaller

Explanation:

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What must the charge (sign and magnitude) of a 3.45 g particle be for it to remain stationary when placed in a downward-directed

Pani-rosa [81]

charge must be equal to 5.74 ×10⁻⁵

In the question it is said that the particle remains stationary which means the the net force on the particle is zero. So, the counterbalancing forces must be equal which means weight is equal to upward electric force.

→ Fnet =0

→ mg = qE

substituting the values we get :

0.00345 × 9.81 = q × 590

→ q = 5.74 ×10⁻⁵

Hence the charge must be equal to 5.74 ×10⁻⁵.

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# SPJ4

8 0

2 years ago

A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal

dsp73

Answer:

$v\approx 8.570\,\frac{m}{s}$

Explanation:

The equation of equlibrium for the box is:

$\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a$

The formula for the acceleration, given in $\frac{m}{s^{2}}$ , is:

$a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}$

Velocity can be derived from the following definition of acceleration:

$a = v\cdot \frac{dv}{dx}$

$v\, dv = a\, dx$

$\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx$

$\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg} \int\limits^{17\,m}_{0\,m}\, dx - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx$

$\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}$

$v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }$

The speed after the box has travelled 17 meters is:

$v\approx 8.570\,\frac{m}{s}$

3 0

3 years ago

The substances referenced in the table are being considered for use in cooking materials. Since the materials must be melted dur

frutty [35]

Brass requires more energy than silver

Aluminum requires more energy than copper

Aluminum requires more energy than brass

Explanation:

The specific heat capacity of a substance indicates the amount of heat energy required to raise 1 kg of that substance by 1 degree in temperature.

Mathematically:

$C=\frac{Q}{m\Delta T}$

where

Q is the heat supplied to the substance

m is the mass of the substance

$\Delta T$ is the change in temperature

Therefore, the higher the specific heat capacity of a substance, the more energy is needed to increase its temperature.

Here we can compare the specific heat capacity of the materials mentioned:

Silver: $C=0.233 J/gK$

Brass: $C=0.380 J/gK$

Platinum: $C=0.130 J/gK$

Aluminium: $C=0.910 J/gK$

Copper: $C=0.390 J/gK$

Therefore, the correct statements are:

Brass requires more energy than silver

Aluminum requires more energy than copper

Aluminum requires more energy than brass

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3 0

3 years ago

A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur

Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0

3 years ago

Why freezer is made in the upper part of refrigrator

svp [43]

<h2>Question</h2>

why freezer is made in the upper part of refrigrator

<h2>✒ Answer</h2>

the cold air produced from it is denser than the warmer air in the bottom

<h3>Explaination</h3>

Freezer is normally provided at the top of the refrigerator, because density of the cold air is high compared to the hot air. In a refrigerator the air contacts with the cooling coil and gets cooling.Because of the high density the cold air gets down and the warm air/hot air moves upward and gets cooling from the cooling coil/evaporator coil. This process is repeated. If the Freezer is provided at the bottom place of the refrigerator, the cold air can't to move full area of the refrigerator. So the freezer is normally provided at the top at the refrigerator

4 0

2 years ago