PE=?, m=.6kg, g=10m/s2, h=35m<br><br> PLS HELP I NEED THIS DONE

Nataly_w [17]

Answer:

9,200 m

Explanation:

1 kilometer = 1000 meters so if there is 9 kilometers then it is 9000 meters, then .2 kilometer is equal to 200 meters. Add the numbers together to get 9,200 meters

5 0

4 years ago

Nathan is standing 2 feet away from a plane mirror. Hiva is standing 5 feet further than

Viktor [21]

Answer:

They are standing 3 feet away from each other

Explanation:

If Nathan Is standing 2 feet away from the mirror. Then Hiva is standing 7 feet away from the mirror is she is standing 5 feet further than him. They are standing 3 feet away from each other.

3 0

3 years ago

In which of these samples do the molecules most likely have the most kinetic energy? (2 points)

Vinvika [58]

D, water vapor. Gaseous state would have more kinetic energy, they are moving faster. If you have to compare the same state, then higher temperature would have the higher kinetic energy. But if you have solid and liquid at the same temperature - then liquid would have more.

4 0

3 years ago

A rock is thrown off a cliff at an angle of 53° with respect to the horizontal. The cliff is 100 m high. The initial speed of th

Nadusha1986 [10]

(a) 129.3 m

The motion of the rock is a projectile motion, consisting of two indipendent motions along the x- direction and the y-direction. In particular, the motion along the x- (horizontal) direction is a uniform motion with constant speed, while the motion along the y- (vertical) direction is an accelerated motion with constant acceleration $g=-9.8 m/s^2$ downward.

The maximum height of the rock is reached when the vertical component of the velocity becomes zero. The vertical velocity at time t is given by

$v(t) = v_0 sin \theta +gt$

where

$v_0 = 30 m/s$ is the initial velocity of the rock

$\theta=53^{\circ}$ is the angle

t is the time

Requiring $v(t)=0$ , we find the time at which the heigth is maximum:

$0=v_0 sin \theta + gt\\t=\frac{-v_0 sin \theta}{g}=-\frac{(30)(sin 53^{\circ})}{(-9.8)}=2.44 s$

The heigth of the rock at time t is given by

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

Where h=100 m is the initial heigth. Substituting t = 2.44 s, we find the maximum height of the rock:

$y=100+(30)(sin 53^{\circ})(2.44)+\frac{1}{2}(-9.8)(2.44)^2=129.3 m$

(b) 44.1 m

For this part of the problem, we just need to consider the horizontal motion of the rock. The horizontal displacement of the rock at time t is given by

$x(t) = (v_0 cos \theta) t$

where

$v_0 cos \theta$ is the horizontal component of the velocity, which remains constant during the entire motion

t is the time

If we substitute

t = 2.44 s

Which is the time at which the rock reaches the maximum height, we find how far the rock has moved at that time:

$x=(30)(cos 53^{\circ})(2.44)=44.1 m$

(c) 7.58 s

For this part, we need to consider the vertical motion again.

We said that the vertical position of the rock at time t is

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

By substituting

y(t)=0

We find the time t at which the rock reaches the heigth y=0, so the time at which the rock reaches the ground:

$0=100+(30)(sin 53^{\circ})t+\frac{1}{2}(-9.8)t^2\\0=100+23.96t-4.9t^2$

which gives two solutions:

t = -2.69 s (negative, we discard it)

t = 7.58 s --> this is our solution

(d) 136.8 m

The range of the rock can be simply calculated by calculating the horizontal distance travelled by the rock when it reaches the ground, so when

t = 7.58 s

Since the horizontal position of the rock is given by

$x(t) = (v_0 cos \theta) t$

Substituting

$v_0 = 30 m/s\\\theta=53^{\circ}$

and t = 7.58 s we find:

$x=(30)(cos 53^{\circ})(7.58)=136.8 m$

(e) (36.1 m, 128.3 m), (72.2 m, 117.4 m), (108.3 m, 67.4 m)

Using the equations of motions along the two directions:

$x(t) = (v_0 cos \theta) t$

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

And substituting the different times, we find:

$x(2.0 s)=(30)(cos 53^{\circ})(2.0)=36.1 m$

$y(2.0 s)= 100+(23.96)(2.0)-4.9(2.0)^2=128.3 m$

$x(4.0 s)=(30)(cos 53^{\circ})(4.0)=72.2 m$

$y(4.0 s)= 100+(23.96)(4.0)-4.9(4.0)^2=117.4 m$

$x(6.0 s)=(30)(cos 53^{\circ})(6.0)=108.3 m$

$y(6.0 s)= 100+(23.96)(6.0)-4.9(6.0)^2=67.4 m$

3 0

3 years ago

How do dominant alleles and recessive alleles differ?

Stells [14]

Answer:

A dominant allele produces a dominant trait in individuals who have one copy of the allele, that can come from one parent. To produces a recessive trait, the child must have two copies of the recessive allele, one from each parent.

Explanation:

The terms dominant and recessive describe the patterns of certain traits. They describe how likely it is for certain traits to pass from parent offspring in humans and animals. The two copies of each gene (alleles), can be slightly different from each other. The differences can cause variations in the protein that’s produced, Proteins affect traits, so variations in protein activity or expression can create different phenotypes.

A dominant allele produces a dominant phenotype (trait) in individuals who have one copy of the allele, which can come from one parent. For a recessive allele to produce a recessive phenotype, the individual must have two copies, one from each parent. A person with one dominant and one recessive allele for a gene will have a dominant phenotype. They are generally considered carriers of the recessive allele- the recessive allele is there, but the recessive phenotype is not.

8 0

3 years ago

PE=?, m=.6kg, g=10m/s2, h=35m PLS HELP I NEED THIS DONE