A dog walks 12 meters to the west and then 16 meters back to the east

A string attached to an airborne kite was maintained at an angle of 40.0 with the ground. if 120m of string was reeled in to ret

AleksAgata [21]

The hypotenuse is measured at 120 meters of string, and you need to solve for the leg of the triangle that is horizontal. The degree is 40, so use trigonometry to figure it out.
Cosin (40) is equal to around .766
Adjacent/Hypotenuse
x/120 = cos40
Answer: 91.92533.
If you use 3 significant figures it should be 91.9 meters.

5 0

3 years ago

A lab cart with a mass of 15 kg is moving with constant velocity, v, along a straight horizontal track. A student drops a 2 kg m

lbvjy [14]

The equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ (option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.

The horizontal momentum is given by:

$p_{i} = p_{f}$

$m_{1}v_{1}_{i} + m_{2}v_{2}_{i} = m_{1}v_{1}_{f} + m_{2}v_{2}_{f}$

Where:

m₁: is the mass of the lab cart = 15 kg

m₂: is the mass of the object dropped = 2 kg

$v_{1}_{i}$ : is the initial velocity of the lab cart

$v_{2}_{i}$ : is the initial velocity of the object = 0 (it is dropped)

$v_{1}_{f}$ : is the final velocity of the lab cart

$v_{2}_{f}$ : is the final velocity of the object

Then, the horizontal momentum is:

$15v_{1}_{i} + 2*0 = 15v_{1}_{f} + 2v_{2}_{f}$

When the object is dropped into the lab cart, the final velocity of the lab cart and the object will be the same, so:

$15v_{1}_{i} + 2*0 = v_{f}(15 + 2)$

Therefore, the equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ represents the horizontal momentum (option 3).

Learn more about linear momentum here:

brainly.com/question/2141713?referrer=searchResults

brainly.com/question/2400186?referrer=searchResults

I hope it helps you!

4 0

2 years ago

An empty graduated cylinder weighs 55.26 g. When filled with 48.1 mL of an unknown liquid, it weighs 92.39 g. The density of the

katrin [286]

<h2>Density of the unknown liquid is 771.93 kg/m³</h2>

Explanation:

An empty graduated cylinder weighs 55.26 g

Weight of empty cylinder = 55.26 g = 0.05526 kg

Volume of liquid filled = 48.1 mL = 48.1 x 10⁻⁶ m³

Weight of cylinder plus liquid = 92.39 g = 0.09239 kg

Weight of liquid = 0.09239 - 0.05526

Weight of liquid = 0.03713 kg

We have

Mass = Volume x Density

0.03713 = 48.1 x 10⁻⁶ x Density

Density = 771.93 kg/m³

Density of the unknown liquid is 771.93 kg/m³

5 0

2 years ago

If you stood on a planet with four times the mass of Earth, and twice Earth's radius, how much would you weigh?

nikdorinn [45]

Answer:

1/4 times your earth's weight

Explanation:

assuming the Mass of earth = M

Radius of earth = R

∴ the mass of the planet= 4M

the radius of the planet = 4R

gravitational force of earth is given as = $\frac{GM}{R^{2} }$

where G is the gravitational constant

Gravitational force of the planet = $\frac{G4M}{(4R)^{2} }$

= $\frac{G4M}{16R^{2} }$

= $\frac{GM}{4R^{2} }$

recall, gravitational force of earth is given as = $\frac{GM}{R^{2} }$

∴Gravitational force of planet = 1/4 times the gravitational force of the earth

you would weigh 1/4 times your earth's weight

3 0

3 years ago

h(t) = - 16t2 + 64t + 112 where t is the time in seconds. After how many seconds does the arrow reach it maximum height? Round t

laila [671]

Answer:

2 seconds

Explanation:

The function of height is given in form of time. For maximum height, we need to use the concept of maxima and minima of differentiation.

$h(t)=-16t^{2}+64t+112$

Differentiate with respect to t on both the sides, we get

$\frac{dh}{dt}=-32t+64$

For maxima and minima, put the value of dh / dt is equal to zero. we get

- 32 t + 64 = 0

t = 2 second

Thus, the arrow reaches at maximum height after 2 seconds.

8 0

2 years ago