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3 years ago

The total energy of a 0.050 kg object travelling at 0.70 c is:

Physics

2 answers:

Elena L [17]3 years ago

8 0

Answer:

1.1025×10^15Joules

No correct option

Explanation:

The type of energy possessed by the object is kinetic energy. Kinetic energy is the energy due to virtue of an object motion.

KE = 1/2MV² where;

M is the mass of the car = 0.05kg

V is the velocity of the car

Since the car is traveling at 0.7c (c is the speed of light)

speed = 0.7c { 0.7(3×10^8)}

Speed = 2.1×10^8

Substituting this values in the formula given we have;

KE = 1/2×0.05×(2.1×10^8)²

KE = 1.1025×10^15Joules

No correct option.

finlep [7]3 years ago

3 0

What is the unit c denotes here

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4 years ago

Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by 0.100 mm.

alekssr [168]

Answer: The angle of diffraction is 0.498°

Explanation:

To calculate the angle of diffraction, we use the equation given by Bragg, which is:

$n\lambda =2d\sin \theta$

where,

n = order of diffraction = 3

$\lambda$ = wavelength of the light = $580nm=5.80\times 10^{-7}m$ (Conversion factor: $1m=10^{9}nm$ )

d = spacing between the crystal planes = 0.100 mm = $1.0\times 10^{-4}$ (Conversion factor: 1 m = 1000 mm)

$\theta$ = angle of diffraction = ?

Putting values in above equation:

$3\times 5.80\times 10^{-7}=2\times 1.00\times 10^{-4}\sin \theta\\\\\sin \theta = 0.0087\\\\\theta=\sin ^{-1} (0.0087)=0.498$

Hence, the angle of diffraction is 0.498°

6 0

3 years ago

Calculate the kinetic energy of a 0.032 kg ball as it leaves a hand to be thrown upwards at 6.2 m/s

AnnZ [28]

Answer:

The ball will have a kinetic energy of 0.615 Joules.

Explanation:

Use the kinetic energy formula

$E_k = \frac{1}{2}mv^2 = \frac{1}{2}0.032kg\cdot 6.2^2 \frac{m^2}{s^2}= 0.615J$

The kinetic energy at the moment of leaving the hand will be 0.615 Joules. (From there on, as it ball is traveling upwards, this energy will be gradually traded off with potential energy until the ball's velocity becomes zero at the apex of the flight)

3 0

3 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago

A 6.8 kg bowling ball and 7.4 kg bowling ball rest on a rack 0.74 m apart. What is the force of gravity pulling each ball toward

zimovet [89]

The gravitational force between the two balls is $6.13\cdot 10^{-9} N$

Explanation:

The magnitude of the gravitational force between two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where :

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

For the balls in this problem, we have

$m_1 = 6.8 kg$

$m_2 = 7.4 kg$

r = 0.74 m

Substituting into the equation, we find the gravitational force between the two balls:

$F=(6.67\cdot 10^{-11})\frac{(6.8)(7.4)}{(0.74)^2}=6.13\cdot 10^{-9}N$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0

3 years ago