Help! I don’t really know what it’s asking

Objects are lighter on the moon than they are on earth. if an object A weighs 25lbs on the Moon and another object B weighs 25 N

solong [7]

Answer:

a. Object A

Explanation:

The mass of an object implies the quantity of matter in it, while the weight is the amount of gravitational force applied on an object.

The object A has a mass of 25 lbs, but object B on the earth has a weight, W, of 25 N.

So that,

For object A on the moon, mass = 25 lbs

For object B on the earth, W = 25 N,

W = m x g

25 = m x 10 (g = 10 m/ $s^{2}$ )

m = $\frac{25}{10}$

= 2.5 lbs

Mass of object B is 2.5 lbs.

Therefore, the mass of the object A is more than that of B.

5 0

2 years ago

A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is

lesantik [10]

Answer:

35870474.30504 m

Explanation:

r = Distance from the surface

T = Time period = 24 h

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

m = Mass of the Earth = 5.98 × 10²⁴ kg

Radius of Earth = $6.38\times 10^6\ m$

The gravitational force will balance the centripetal force

$\dfrac{GMm}{R^2}=m\dfrac{v^2}{R}\\\Rightarrow v=\sqrt{\dfrac{GM}{R}}$

$T=\dfrac{2\pi r}{v}\\\Rightarrow T=\dfrac{2\pi r}{\sqrt{\dfrac{GM}{r}}}$

From Kepler's law we have relation

$T^2=\dfrac{4\pi^2r^3}{GM}\\\Rightarrow r^3=\dfrac{T^2GM}{4\pi^2}\\\Rightarrow r=\left(\dfrac{(24\times 3600)^2\times 6.67\times 10^{-11}\times 5.98\times 10^{24}}{4\pi^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=42250474.30504\ m$

Distance from the center of the Earth would be

$42250474.30504-6.38\times 10^6=\mathbf{35870474.30504\ m}$

8 0

3 years ago

A 0.4-kg toy train car moving forward at 3 m/s collides with and sticks to a 0.8–kg toy car that is traveling in the opposite di

Gemiola [76]

Hey there!

Seems like you're looking for the size and direction to the final velocity of the two cars. To find it, you must solve it like this.

0.4 kg(3 m/s) + 0.8kg(–2 m/s) = 1.2 kg m/s -1.6 kg m/s = –0.4 kg m/s

–0.4 kg m/s = 1.2 kg(v) = (–0.4 kg m/s)/(1.2 kg) = v = –0.33 m/s

So, the cars are traveling at -0.33 m/s in the direction of the second car.

Hope this helps

<em>Tobey</em>

4 0

2 years ago

An artillery shell is launched on a flat, horizontal field at an angle of α = 40.8° with respect to the horizontal and with an i

Lina20 [59]

Answer:

1317.4 m

Explanation:

We are given that

Angle= $\alpha=40.8^{\circ}$

Initial speed = $v_0=346m/s$

We have to find the horizontal distance covered by the shell after 5.03 s.

Horizontal component of initial speed= $v_{ox}=v_0cos\theta=346cos40.8=261.9m/s$

Vertical component of initial speed= $v_{oy}=346sin40.8=226.1m/s$

Time=t=5.03 s

Horizontal distance = $Horizontal\;velocity\times time$

Using the formula

Horizontal distance= $261.9\times 5.03$

Horizontal distance=1317.4 m

Hence, the horizontal distance covered by the shell=1317.4 m

8 0

2 years ago

Keeping the mass at 1.0 kg and the velocity at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius

Paha777 [63]

Answer:

The centripetal acceleration for the first radius; 2.0 m = 50 m/s²

The centripetal acceleration for the second radius; 4.0 m = 25 m/s²

The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²

The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

Explanation:

Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

$a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2$

The centripetal acceleration for the second radius; 4.0 m

$a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2$

The centripetal acceleration for the third radius; 6.0 m

$a_c_3= \frac{(10)^2}{6} \\\\a_c_3= 16.67 \ m/s^2$

The centripetal acceleration for the fourth radius; 8.0 m

$a_c_4= \frac{(10)^2}{8} \\\\a_c_4= 12.5 \ m/s^2$

The centripetal acceleration for the fifth radius; 10.0 m

$a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2$

6 0

3 years ago