I need help the option for D is the golfball isnt accelerating

A machine can never be 100% efficient because some work is always lost due to .

klasskru [66]

Friction
Hope it helped

5 0

4 years ago

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Find the y-component of this vector:

NARA [144]

Answer:

Dy = 49.01 [m]

Explanation:

The vertical component of the vector can be determined with the sine of the angle.

Dy = 92.5*sin(32)

Dy = 49.01 [m]

6 0

3 years ago

If a planet has 3 times the radius of the Earth, but has the same density as the Earth, what is the gravitational acceleration a

xz_007 [3.2K]

Answer:

The Gravitational Acceleration of the Big planet is G = 3g

where G is the gravitational acceleration of the Big Planet and g is the gravitational acceleration of the Earth

So, it becomes G = 3 x 9.8 $ms^{-2}$

G = 29.4 $ms^{-2}$

Explanation:

Let's start by calculating the Density of the Earth of mass m anad Radius r

We know that mass is density times the volume

ρ = $\frac{m}{V}$

m = ρV

we know the volume is V = $\frac{4}{3}$ $πr^{3}$ (please ignore the symbol of Pi)

m = ρ $\frac{4}{3}$ $πr^{3}$

Calculating the Density of Big Planet

→For Big Planet we know that radius is 3-times the radius of Earth, that is 3r

M = ρ $\frac{108}{3}$ $π(r)^{3}$

So in conclusion, the mass of Earth and the mass of Big planet are related as M = 27m

Now let's come to the Gravitational Force, we know that gravitational force is directly proportional to the mass of the body and inversely proportional to the radius.

→Suppose for Earth: Mass = m, Radius = r

For Big Planet: Mass = M, Radius = R

$\frac{m}{r^{2} }$ : $\frac{M}{R^{2} }$ = g : G

$\frac{Gm}{r^{2} }$ = $\frac{gM}{R^{2} }$

G = $\frac{gMr^{2}}{mR^{2}}$

→ Putting the value of R = 3r and M = 27m

G = $\frac{g27mr^{2}}{m(3r)^{2}}$

By cutting the terms, we get

G = 3g

value of g= 9.8 $ms^{-2}$

G = 3 x $9.8ms^{-2}$

G = 29.4 $ms^{-2}$

5 0

4 years ago

A 25.0 g marble sliding to the right at 20.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same dir

ikadub [295]

In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

KE1 = KE2

The kinetic energy of the system before the collision is solved below.

KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²

This value should also be equal to KE2, which can be calculated using the conditions after the collision.

KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)

The value of x from the equation is 17.16 cm/s.

Hence, the answer is 17.16 cm/s.

6 0

4 years ago

What can be concluded about the spread of the histogram? A histogram titled Maria's Monthly Jogging has miles run on the x-axis

Aleonysh [2.5K]

Answer:

I think it is A

Explanation:

l

8 0

3 years ago

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