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3 years ago

Dave throws a 10 kg bowling ball straight up in the air. At the very tippy-top of its path, what is it's momentum?

Physics

1 answer:

svetoff [14.1K]3 years ago

8 0

Answer:Zero

Explanation:

Given

mass of ball $m=10\ kg$

If the ball is thrown upward then at maximum point velocity of ball is zero because ball is no longer able to move upward

Momentum(P) of a particle is given by

$P=mass\times velocity$

$P=10\times 0$

$P=0$

Therefore at the highest point momentum is zero .

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2. Light waves of the wavelength of 650 nm and 500 nm produce interference fringes on a screen at a distance of 1m from a double

satela [25.4K]

Answer:

least distance= 13mm

ratio of the lattice = 1 : 0.71 : 0.58

Explanation:

given λ₁ = 650nm = 650×10⁻⁹m, λ₂ = 500nm = 500×10⁻⁹m

5 0

3 years ago

if an object is being acted on by two forces a push and a pull of 6N what is the net force of the object

Paul [167]

The concepts of force, mass, and weight play critical roles. Newton's Laws of. Motion ... the person stops pushing? ... F net =10 N 2 N. = 8 N (to the right) a = F net m. = 8 N. 5 kg. =1.6 m s. 2 ... Two equal forces act on an object in the directions shown. If these ... Two connected carts being accelerated by a force F applied by.

6 0

3 years ago

Suppose that a comet that was seen in 563 A.D. by Chinese astronomers was spotted again in year 1951. Assume the time between ob

Mars2501 [29]

Answer:

$a=2.77*10^{13}m$

$R_a=5.49*10^{13}m$

Explanation:

The period of the comet is the time it takes to do a complete orbit:

T=1951-(-563)=2514 years

writen in seconds:

$2514years*\frac{3,154*10^7s}{1year}=7.93 *10^{10}s$

Since the eccentricity is greater than 0 but lower than 1 you can know that the trajectory is an ellipse.

Therefore, if the mass of the sun is aprox. 1.99e30 kg, and you assume it to be much larger than the mass of the comet, you can use Kepler's law of periods to calculate the semimajor axis:

$T^2=\frac{4\pi^2}{Gm_{sun}}a^3\\ a=\sqrt[3]{\frac{Gm_{sun}T^2}{4\pi^2} } \\a=1.50*10^{6}m$

Then, using the law of orbits, you can calculate the greatest distance from the sun, which is called aphelion:

$R_a=a(1+e)\\R_a=2.77*10^{13}(1.986)\\R_a=5.49*10^{13}m$

8 0

3 years ago

To pull a 57 kg crate across a horizontal frictionless floor, a worker applies a force of 230 N, directed 34° above the horizont

lisabon 2012 [21]

Answer:

$a. W_w=235\ J\\b. W_g=-343.54\ J\\c. F_N=463.100\ N\\d. W_t=235\ J$

Explanation:

Given: that,

Angle of inclination of the surface, $\theta=34^{\circ}$

mass of the crate, $m=57\ kg$

Force applied along the surface, $F=230\ N$

distance the crate moves after the application of force, $s=1.1\ m$

a) work done = F× s

work done = 230 × 1.1

work done = 253 J

b) Work done by the gravitational force:

$W_g=m.g\times h$

where:

g = acceleration due to gravity

h = the vertically downward displacement

Now, we find the height:

$h=s\times sin\ \thetah=1.1\times sin\ 34^{\circ}h=0.615\ m$

So, the work done by the gravity:

$W_g=57\times 9.8\times (-0.615) \\= - 343.54 J$

∵direction of force and displacement are opposite.

= - 343.54J

The normal reaction force on the crate by the inclined surface:

$F_N=m.g.cos\ \thetaF_N=57\times 9.8\times cos\ 34F_N=463.100\ N$

Total work done on crate is with respect to the worker:

$W_t=235\ J$

5 0

3 years ago

A "home-made" solid propellant rocket has an initial mass of 9 kg; 6.8 kg of this is fuel. The rocket is directed vertically upw

nlexa [21]

Answer:

v = 1176.23 m/s

y = 741192.997 m = 741.19 km

Explanation:

Given

M₀ = 9 Kg (Initial mass)

me = 0.225 Kg/s (Rate of fuel consumption)

ve = 1980 m/s (Exhaust velocity relative to rocket, leaving at atmospheric pressure)

v = ? if t = 20 s

y = ?

We use the equation

v = ∫((ve*me)/(M₀ - me*t)) dt - ∫g dt where t ∈ (0, t)

⇒ v = - ve*Ln ((M₀ - me*t)/M₀) - g*t

then we have

v = - 1980 m/s*Ln ((9 Kg - 0.225 Kg/s*20 s)/(9 Kg)) - (9.81 m/s²)(20 s)

v = 1176.23 m/s

then we apply the formula

y = ∫v dt = ∫(- ve*Ln ((M₀ - me*t)/M₀) - g*t) dt

⇒ y = - ve* ∫ Ln ((M₀ - me*t)/M₀) dt - g*∫t dt

⇒ y = - ve*(Ln((M₀ - me*t)/M₀)*t + (M₀/me)*(M₀ - me*t - M₀*Ln(M₀ - me*t))) - (g*t²/2)

For t = 20 s we have

y = Ln((9 Kg - 0.225 Kg/s*20 s)/9 Kg)*(20 s) + (9 Kg/0.225 Kg/s)*(9 Kg - 0.225 Kg/s*20 s - 9 Kg*Ln(9 Kg - 0.225 Kg/s*20 s)) - (9.81 m/s²*(20 s)²/2)

⇒ y = 741192.997 m = 741.19 km

The graphs are shown in the pics.

6 0

3 years ago