Ask question

Ask question

All categories

1 year ago

14

When were Earth’s landmasses first recognizable as the continents we know today? 10 million years ago 135 million years ago 180

million years ago 300 million years ago

2 answers:

Bess [88]1 year ago

6 0

Answer:

b

Explanation:

i took the test

cluponka [151]1 year ago

6 0

Answer:

b 135 million years

Explanation:

You might be interested in

A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then

Elan Coil [88]

Answer:

A) $W_{ff} =-744.12J$

B) $F_f=-W_{ff}*sin\theta /hy = 112.75N$

C) $F_{f2}=207.58N$

Explanation:

This question is incomplete. The full question was:

<em>A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s. </em>

<em>Part (a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement. </em>

<em>Part (b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder. </em>

<em>Part (c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.</em>

For part A), we make a balance of energy to calculate the work done by the friction force:

$W_{ff}=\Delta E$

$W_{ff}=1/2*m*vf^2-m*g*hy$

$W_{ff}=-744.12J$

For part B), we use our previous value for the work:

$W_{ff}=-F_f*(hy/sin\theta)$ Solving for friction force:

$F_f=-W_{ff}*sin\theta /hy$

$F_f=112.75N$

For part C), we first calculate the acceleration by kinematics and then calculate the module of friction force by dynamics:

$Vf^2=Vo^2+2*a*d$

Solving for a:

$a=-3.844m/s^2$

Now, by dynamics:

$|F_f|=|m*a|$

$|F_f|=207.58N$

8 0

2 years ago

Two workhorses tow a barge along a straight canal. Each horse exerts a constant force of magnitude F, and the tow ropes make an

morpeh [17]

Answer:

<em>a) Fvt cosθ</em>

<em>b) Fv cosθ</em>

<em></em>

Explanation:

Each horse exerts a force = F

the rope is inclined at an angle = θ

speed of each horse = v

a) In time t, the distance traveled d = speed x time

i.e d = v x t = vt

also, the resultant force = F cosθ

Work done W = force x distance

W = F cosθ x vt = <em>Fvt cosθ</em>

<em></em>

b) Power provided by the horse P = force x speed

P = F cosθ x v

P = <em>Fv cosθ</em>

8 0

1 year ago

A cement factory emits 900 kilograms of CO2 to produce 1,000 kilograms of cement. A fully grown tree removes six kilograms of CO

Dmitriy789 [7]

<h3><u>Answer;</u></h3>

To make the factory carbon neutral, the owners need to grow<em><u> 15000 trees</u></em> over an area of land measuring <em><u>75 acres</u></em>

<h3><u>Explanation and solution;</u></h3>

From the information;

900 kg CO2 = 1000 kg Cement

1 tree = 6 kg CO2

1 acre = 200 trees

<em>100000 kg Cement will require;</em>

<em>=(900 × 100000)/1000</em>

<em>= 90,000 kg of CO2</em>

<em>But 1 tree = 6 kg of CO2</em>

<em>Number of trees = 90,000/6</em>

<em> = 15,000 trees </em>

<em>But, 1 acre = 200 trees</em>

<em>Number of acres = 15,000/200</em>

<em> = 75 acres of land </em>

Therefore;

To make the factory carbon neutral, the owners need to grow<em><u> 15000 trees</u></em> over an area of land measuring <em><u>75 acres</u></em>

6 0

2 years ago

Read 2 more answers

For what value m of the clockwise couple will the horizontal component ax of the pin reaction at a be zero? if a couple of that

sergejj [24]

Thank you for posting your question here at brainly. Below is the answer:

sum of Mc = 0 = -Ay(4.2 + 3cos(59)) + (275)(2.1 + 3cos(59)) + M
<span>- Ay = (M + (275*(2.1 + 3cos(59)))/(4.2 + 3cos(59)) </span>

<span>sum of Ma = 0 = (-275)(2.1) - Cy(4.2 + 3cos(59)) + M </span>
<span>- Cy = (M - (275*2.1))/(4.2 + 3cos(59)) </span>

<span>Ay + Cy = 275 = ((M+1002.41)+(M-577.5))/(5.745) </span>
<span>= (2M + 424.91)/(5.745) </span>

<span>M = ((275*5.745) - 424.91)/2 </span>
<span>= 577.483 which rounds off to 577 </span>

<span>Is it maybe supposed to be Ay - Cy = 275</span>

8 0

2 years ago

En la etiqueta de un bote de fabada de 350 g, leemos que su aporte energético es de 1630 kj por cada 100 g de producto a) La can

fredd [130]

Answer:

(a) 153.37 g

(b) 5705 kJ

Explanation:

(a) To find the amount of bean needed by a man you first calculate the equivalence in beans to 2500kJ

$2500kJ*\frac{100g}{1630kJ}=153.37\ g$

Thus, 153.37 g has the energy needed by a man that needs 200kJ per day.

(b) The amount of energy per pot of bean is given by:

$E=350g*\frac{1630kJ}{100g}\\\\E=5705\ kJ$

Thus, the energy is 5705kJ

6 0

2 years ago