The following is an example of which type of reaction?<br> CH3OH + O2 → CO2 + 2H2O + heat

How long does it take for an 8 kg pumpkin to hit the ground if dropped from a height of 55 m.

Sati [7]

Answer:

t = 3.35 s

Explanation:

It is given that,

Mass of a pumpkin, m = 8 kg

It is dropped from a height of 55 m

We need to find the time taken by it to hit the ground.

Initial velocity of the pumpkin, u = 0

Using second equation of motion to find it as follows :

$h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2(55)}{9.8}} \\\\t=3.35\ s$

So, it will take 3.35 seconds to hit the ground.

7 0

3 years ago

LOOKING FOR ALL YOU SCIENTISTS!!

tensa zangetsu [6.8K]

Answer:

Make sure you're getting paid. Unpaid interships are a complete and total waste.

Explanation:

7 0

3 years ago

Read 2 more answers

If a car is traveling at an average speed of 60 kilometers per hour how long does it take to travel 12 kilometers

lara31 [8.8K]

Answer:

The time taken to travel is, t = 12 minutes

Explanation:

Given data,

The speed of the car, v = 60 km/h

The distance of travel, d = 12 km

The time taken for the travel is t = ?

The speed is defined as the distance divided by the time taken to travel. The formula for speed is,

v = d/t

∴ t = d/v

t = 12 km / 60 km/h

t = 0.2 h

t = 12 minutes

Hence, the time taken to travel is, t = 12 minutes.

5 0

3 years ago

A puck sliding on ice approaches a 20 degrees ramp at 10 m/s with coefficient of friction 0.2. How high will it go up the ramp a

timama [110]

Answer:

it will go up along the inclined plane by d = 9.62 m

Explanation:

As we know that puck is moving upward along the slide

then the net force opposite to its speed is given as

$F_{net} = - mgsin\theta - \mu mgcos\theta$

so here deceleration is given as

$a = -g(sin\theta + \mu cos\theta)$

now plug in all values in it

$a = -9.81(sin20 + 0.2cos20)$

$a = -5.2 m/s^2$

now the distance covered by the puck along the plane is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - 10^2 = 2(-5.2)d$

$d = 9.62 m$

3 0

4 years ago

0.45 kg soccer ball changes its velocity by 20.0 m/s due to a force applied to it in 0.10 seconds. What force was necessary for

Paladinen [302]

Assuming the accleration applied was constant, we have

$v=v_0+at\implies v_0+20.0\,\dfrac{\mathrm m}{\mathrm s}=v_0+a(0.10\,\mathrm s)$

$\implies20.0\,\dfrac{\mathrm m}{\mathrm s}=a(0.10\,\mathrm s)$

$\implies a=200\,\dfrac{\mathrm m}{\mathrm s^2}$

Then the force applied to the ball is given by

$F=ma=(0.45\,\mathrm{kg})\left(200\,\dfrac{\mathrm m}{\mathrm s^2}\right)$

$\implies F=90\,\dfrac{\mathrm{kg}\,\mathrm m}{\mathrm s^2}=90.\,\mathrm N$

8 0

3 years ago

The following is an example of which type of reaction? CH3OH + O2 → CO2 + 2H2O + heat