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2 years ago

true or false:acceleration toward the center of a curved or circular path is called gravitational acceleration.

1 answer:

5 0

Nope. It's called 'centripetal' acceleration. The force that created it MAY be gravitational, but it doesn't have to be. For things on the surface of the Earth moving in circles, it's never gravity.

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a 1.50*10^-5 C charge feels a 2.89*10^-3 N force when it moves 288m/s perpendicular (90) deg to a magnetic field. how strong is

Sunny_sXe [5.5K]

6.68, -1
Explanation: correct for acellus

5 0

2 years ago

A block spring system oscillates on a frictionless surface with an amplitude of 10\text{ cm}10 cm and has an energy of 2.5 \text

antoniya [11.8K]

Answer:

The energy of the system is 15 J.

Explanation:

Given that,

Energy E = 2.5 J

Amplitude = 10 cm

We need to calculate the spring constant

Using formula of mechanical energy of the system

$E=\dfrac{1}{2}kA^2$

Put the value into the formula

$2.5=\dfrac{1}{2}k\times(10\times10^{-2})^2$

$k=\dfrac{2.5\times2}{(10\times10^{-2})^2}$

$k=500\ N/m$

If the block is replaced by a block with twice the mass of the original block

Amplitude = 6 cm

We need to calculate the energy

Using formula of mechanical energy

$E=\dfrac{1}{2}kA^2$

Put the value into the formula

$E=\dfrac{1}{2}\times500\times(6\times10^{-2})$

$E=15\ J$

Hence, The energy of the system is 15 J.

8 0

2 years ago

An air-conditioning system is to be filled from a rigid container that initially contains 5 kg of liquid R-134a at 24°C. The val

OverLord2011 [107]

Answer:

Answer

The Final Quality of teh R-134a in the container is 0.5056

The Total Heat transfer is $Q_{in} = 22.62 KJ$

Explanation:

Explanation is in the following attachments

3 0

2 years ago

If a metallic wire of cross sectional area 3.0 ´ 10-6 m2 carries a current of 6.0 A and has a mobile charge density of 4.24 ´ 10

elena-14-01-66 [18.8K]

Answer:

The drift velocity is $v_d=2.9\times 10^{-4}\ m/s.$

Explanation:

Given :

Area of metallic wire, $A = 3\times 10^{-6}\ m^2$ .

Current through wire , $I=6 \ A.$

Mobile charge density , $n=4.24\times 10^{28} \ carriers/m^3.$

Charge value , $e=1.6\times 10^{-19}\ C.$

We need to find drift velocity , $v_d.$

Now, we know :

$I=neAv_d$

Therefore, $v_d=\dfrac{I}{neA}$

Putting all given values in above equation we get,

$v_d=\dfrac{6}{4.24\times 10^{28}\times 1.6\times 10^{-19} \times 3 \times 10^{-6}}$

$v_d=2.9\times 10^{-4}\ m/s.$

Hence, this is the required solution.

8 0

2 years ago

The mass of the Sun is 2multiply1030 kg, and the mass of the Earth is 6multiply1024 kg. The distance from the Sun to the Earth i

spayn [35]

Answer:

a) 3.56 x 10^22 N

b) 3.56 x 10^22 N

Explanation:

Mass of the sun M = 2 x 10^30 kg

mass of the Earth m = 6 x 10^24 kg

Distance between the sun and the Earth R = 1.5 x 10^11 m

From Newton's law,

F = $\frac{GMm}{R^2}$

where F is the gravitational force between the sun and the Earth

G is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2

m is the mass of the Earth

M is the mass of the sun

R is the distance between the sun and the Earth.

Substituting values, we have

F = $\frac{6.67*10^{-11}*2*10^{30}*6*10^{24}}{(1.5*10^{11})^2}$ = 3.56 x 10^22 N

A) The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to 3.56 x 10^22 N

b) The force exerted by the Earth on the Sun = 3.56 x 10^22 N

7 0

2 years ago