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2 years ago

Two gravitational forces act on a

Physics

1 answer:

Alik [6]2 years ago

3 0

Answer:

Explanation:

Depends on the location of the two forces. If they are aligned on the same side of the object, you would simply add.

X -----------F1 -------F2

X is the object. F1 and F2 are both masses which create a gravitational force. They both are the form of Fx = G * m1 * m2 / r^2. The total force is F1 + F2

If they are are on either side of the object, you subtract.

F1 ---------X ---------F2

Fx = F1 - F2

Any other location of F1 and F2 is much more complicated by the use of trigonometry.

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A certain shade of blue has a frequency of 7.24 × 1014 Hz. What is the energy of exactly one photon of this light?

ANEK [815]

E = hf

E = 6.63* 10 ⁻³⁴ * 7.24* 10¹⁴

<span>E = 4.80012 × 10⁻¹⁹ J</span>

4 0

3 years ago

Read 2 more answers

g A change in the initial _____ of a projectile changes the range and maximum height of the projectile.

docker41 [41]

Answer:

Velocity.

Explanation:

Projectile motion is characterized as the motion that an object undergoes when it is thrown into the air and it is only exposed to acceleration due to gravity.

As per the question, 'any change in the initial velocity of the projectile(object having gravity as the only force) would lead to a change in the range as well as the maximum height of the projectile.' To illustrate numerically:

Horizontal range: As per expression:

R= ( $u^{2}$ *sin2θ)/g

the range depending on the square of the initial velocity.

Maximum height: As per expression:

H= ( $u^{2}$ * $sin^{2}$ θ )/2g

the maximum distance also depends upon square of the initial velocity.

7 0

3 years ago

A mouse jumps horizontally from a box of height 0.25m. If the mouse jumps with a speed of 2.1 m/s, how far from the box does th

Sladkaya [172]

The mouse would land 0.47 m away from the box.

3 0

3 years ago

Read 2 more answers

The distance between the earth and sun is 1.5 x 108 kilometers and the speed of light is 3.00 x 108 meters per second. Calculate

butalik [34]

Answer:

time = 8.3333 minutes.

Explanation:

distance between earth and sun = 1.5 * $10^{8}km$

speed of light = 3* $10^{8}m/s$

convert the distance unit from km to m so we can have uniform units.

distance between earth and sun = 1.5 * $10^{8}*1000$ m

distance between earth and sun = 1.5 * $10^{11}$ m

speed = distance /time

time = distance / speed

time = $\frac{1.5*10^{11} }{3*10^{8} }$

$= 0.5*10^{3}$

time =500 sec

time = 500/60 minutes

time = 8.3333 minutes.

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