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3 years ago

12

Which word is most closely associated with uniformitarianism

1 answer:

stellarik [79]3 years ago

5 0

Can you give me the choices.<span />

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50 points Answer fast please

erma4kov [3.2K]

Answer:

22 hours

Explanation:

6 0

3 years ago

19) A child on a sled starts from rest at the top of a 15.0° slope. If the trip to the bottom takes 15.2 s,

sveticcg [70]

Answer: 288.8 m

Explanation:

We have the following data:

$t=15.2 s$ is the time it takes to the child to reach the bottom of the slope

$V_{o}=0$ is the initial velocity (the child started from rest)

$\theta=15\°$ is the angle of the slope

$d$ is the length of the slope

Now, the Force exerted on the sled along the ramp is:

$F=ma$ (1)

Where $m$ is the mass of the sled and $a$ its acceleration

In addition, if we draw a free body diagram of this sled, the force along the ramp will be:

$F=mg sin \theta$ (2)

Where $g=9.8 m/s^{2}$ is the acceleration due gravity

Then:

$ma=mg sin \theta$ (3)

Finding $a$ :

$a=g sin \theta$ (4)

$a=9.8 m/s^{2} sin(15\°)$ (5)

$a=2.5 m/s^{2}$ (6)

Now, we will use the following kinematic equations to find $d$ :

$V=V_{o}+at$ (7)

$V^{2}=V_{o}^{2}+2ad$ (8)

Where $V$ is the final velocity

Finding $V$ from (7):

$V=at=(2.5 m/s^{2})(15.2 s)$ (9)

$V=38 m/s$ (10)

Substituting (10) in (8):

$(38 m/s)^{2}=2(2.5 m/s^{2})d$ (11)

Finding $d$ :

$d=288.8 m$

6 0

3 years ago

Can someone please help, ;w;?

SOVA2 [1]

I believe the answer is 3). The cell wall provides protection, it doesn’t control movements of materials in and out of the cell.

7 0

3 years ago

Read 2 more answers

What force is required to cause a 5 kg bowling ball to accelerate at 4 m/s2

Scorpion4ik [409]

Answer:

20 N

Explanation:

By Newton's 2nd law,

The rate of change of momentum is directly proportional to the unbalance force applied on the object,

By that you can get the equation,

F = ma

= 5 × 4 = 20 N

4 0

3 years ago

A punter drops a 2.0 kg ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at

pav-90 [236]

Answer:

Explanation:

Given

mass of drop $m=2 kg$

height of fall $h=1 m$

ball leaves the foot with a speed of 18 m/s at an angle of $55^{\circ}$

Velocity of ball just before the collision with the floor

$u^=2gh$

$u=\sqrt{2gh}$

$u=\sqrt{2\times 9.8\times 1}=4.42 m/s$

Impulse delivered in Y direction

$J_y=m(v\sin (55)-(-u))$

$J_y=2(18\sin (55)+4.42)$

$J_y=38.32 kg-m/s$

Impulse in x direction

$J_x=m\times v\cos (55)$

$J_x=2\times 4.42\cos (55)=20.646$

$J_{net}=\sqrt{J_x^2+J_y^2}$

$J_{net}=\sqrt{(38.32)^2+(20.64)^2}$

$J_{net}=43.52 kg-m/s$

at an angle of $\tan \phi =\frac{J_y}{J_x}=\frac{38.32}{20.64}$

$\phi =tan^{-1}(1.856)$

$\phi =61.7^{\circ}$

7 0

3 years ago