We can predict the path of a hurricane because of the work of which of the following scientists?

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f

Alexandra [31]

The acorn was at a height of 4.15 m from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

$s=ut+ \frac{1}{2} at^2$

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

$s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s$

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height 4.15m from the ground before dropping down.

3 0

3 years ago

Identify the sentence that clearly states its meaning.

netineya [11]

d) While taking a shower, I saw a little mouse scurry across the bathroom floor.

This sentence is ordered properly.

7 0

3 years ago

A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit

Lubov Fominskaja [6]

Answer:

58515.9 m/s

Explanation:

We are given that

$d_1=4.7\times 10^{10} m$

$v_i=9.5\times 10^4 m/s$

$d_2=6\times 10^{12} m$

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

$K_i+U_i=K_f+U_f$

$\frac{1}{2}mv^2_i-\frac{GmM}{d_1}=\frac{1}{2}mv^2_f-\frac{GmM}{d_2}$

$\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2}=\frac{1}{2}v^2_f$

$v^2_f=2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})$

$v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}$

Using the formula

$v_f=\sqrt{v^2_i+2GM(\frac{1}{d_2}-\frac{1}{d_1})}$

$v_f=\sqrt{(9.5\times 10^4)^2+2\times 6.7\times 10^{-11}\times 1.98\times 10^{30}(\frac{1}{6\times 10^{12}}-\frac{1}{4.7\times 10^{10})}$

Where mass of sun= $M=1.98\times 10^{30} kg$

$G=6.7\times 10^{-11}$

$v_f=58515.9 m/s$

4 0

3 years ago

A 0.07-kg lead bullet traveling 258 m/s strikes an armor plate and comes to a stop. If all of the bullet's energy is converted t

Lubov Fominskaja [6]

Answer:

temperature change is 262.06°K

Explanation:

given data

mass = 0.07 kg

velocity = 258 m/s

to find out

what is its temperature change

solution

we know here

heat change Q is is equal to kinetic energy that is

KE = 0.5 × m× v² ...........1

here m is mass and v is velocity

KE = 0.5 × 0.07 × 258²

KE = 2329.74 J

and we know

Q = mC∆t .................2

here m is mass and ∆t is change in temperature and C is 127J/kg-K

so put here all value

2329.74 = 0.07 × 127 × ∆t

∆t = 262.06

so temperature change is 262.06°K

7 0

3 years ago

A weight is hung from the ceiling of an elevator by a massless string. Under which circumstances will the tension in the cord be

Oduvanchick [21]

When the elevator is going up (assuming the elevator is acceleration)

When the elevator is accelerating downwards, the total gravitational force would be larger.

If the elevator is accelerating upwards, then the gravitation force would be smaller, thus the tension in the string would be smaller.

4 0

3 years ago