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3 years ago

In physics what does 7.56 × 5.746 equal ?

Physics

1 answer:

lapo4ka [179]3 years ago

7 0

Answer:

43.43

Explanation:

5.746 x 7.56 = 43.43976

As the least number of desimal is two so our awnser should contain two digits after the decimal point.

Ans: 43.43.

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For a 50 kV anode voltage, what is the maximum photon energy of the x-ray radiation?

V125BC [204]

Answer:

The energy of photon, $E=8\times 10^{-15}\ J$

Explanation:

It is given that,

Voltage of anode, $V=50\ kV=50\times 10^3\ V=5\times 10^4\ V$

We need to find the maximum energy of the photon of the x- ray radiation. The energy required to raise an electron through one volt is called electron volt.

$E=eV$

e is charge of electron

$E=1.6\times 10^{-19}\times 5\times 10^4$

$E=8\times 10^{-15}\ J$

So, the maximum energy of the x- ray radiation is $8\times 10^{-15}\ J$ . Hence, this is the required solution.

6 0

3 years ago

Maggie completed a 10000-m race at an average speed of 160

Gala2k [10]

Answer: 200m/min

Explanation:

Divide 10000m by 160m/min, you will get the answer 62.5. You then subtract 12.5 from 62.5 to understand what you will need your answer for the other person’s speed will be. 10000m divided by 50min is 200m/min.

3 0

3 years ago

n deep space, sphere A of mass 47 kg is located at the origin of an x axis and sphere B of mass 110 kg is located on the axis at

Volgvan

Answer:

a)-1.014x $10^{-7$ J

b)3.296 x $10^{-7$ J

Explanation:

For Sphere A:

mass 'Ma'= 47kg

xa= 0

For sphere B:

mass 'Mb'= 110kg

xb=3.4m

a)the gravitational potential energy is given by

$U_{i$ = -GMaMb/ d

$U_{i$ = - 6.67 x $10^{-11}$ x 47 x 110/ 3.4 => -1.014x $10^{-7$ J

b) at d= 0.8m (3.4-2.6) and $U_{i$ =-1.014x $10^{-7$ J

The sum of potential and kinetic energies must be conserved as the energy is conserved.

$K_{i$ + $U_{i$ = $K_{f$ + $U_{f$

As sphere starts from rest and sphere A is fixed at its place, therefore $K_{i$ is zero

$U_{i$ = $K_{f$ + $U_{f$

The final potential energy is

$U_{f$ = - GMaMb/d

Solving for ' $K_{f$ '

$K_{f$ = $U_{i$ + GMaMb/d => -1.014x $10^{-7$ + 6.67 x $10^{-11}$ x 47 x 110/ 0.8

$K_{f$ = 3.296 x $10^{-7$ J

6 0

2 years ago

When a boat moves through the water, the waves in front of the boat bunch up, while the waves behind the boat spread out. this i

liberstina [14]

Answer: Doppler Effect

Doppler Effect can be described as the change in wavelength of a wave like upward shift in frequency for an object whom is approaching and an apparent downward shift in frequency for observers from whom the source is receding. This effect can be observed when a boat moves through the water then the waves in front bunch up while the waves behind the boat spread out.

8 0

3 years ago

A child whirls a 3.00 kg ball on a string .50 m from the axis of rotation in a horizontal circle. The ball makes 1 revolution in

melamori03 [73]

Answers:

a) $0.5 m/s^{2}$

b) $1.5 N$

Explanation:

a) The centripetal acceleration $a_{c}$ of an object moving in a uniform circular motion is given by the following equation:

$a_{c}=\omega^{2} r$

Where:

$\omega=1 \frac{rev}{s}$ is the angular velocity of the ball

$r=0.5 m$ is the radius of the circular motion, which is equal to the length of the string

Then:

$a_{c}=(1 \frac{rev}{s})^{2} 0.5 m$

$a_{c}=0.5 m/s^{2}$ This is the centripetal acceleration of the ball

b) On the other hand, in this circular motion there is a force (centripetal force $F$ ) that is directed towards the center and is equal to the tension ( $T$ ) in the string:

$F=T=m. a_{c}$

Where $m=3 kg$ is the mass of the ball

Hence:

$T=(3 kg)(0.5 m/s^{2})$

$T=1.5 N$ This is the tension in the string

7 0

3 years ago