A weightlifter can lift 220lbs. (assume 1kg weighs 2.2lbs) If he lifts the weight 2m above the floor, how much work did he do?

A bag of cement has a mass of 62 g. What is the mass of the bag of cement in S.I. units (kg)?

NNADVOKAT [17]

The mass of this bag of cement in S.I. units (kg) is equal to 0.062 kilograms.

<u>Given the following data:</u>

Mass of cement = 62 grams.

To calculate the mass of this bag of cement in S.I. units (kg):

<h3>How to convert to S.I. units.</h3>

In Science, kilograms (kg) is the standard unit of measurement or S.I. units of the mass of a physical object. Thus, we would convert the value of the mass of this bag of cement in grams to kilograms (kg) as follows:

<u>Conversion:</u>

1000 grams = 1 kilograms.

62 grams = X kilograms.

Cross-multiplying, we have:

X = $\frac{62}{1000}$

X = 0.062 kilograms.

Read more on mass here: brainly.com/question/13833323

8 0

2 years ago

Select the three correct parts that make up the cell theory.

Lana71 [14]

Answer:

Options A, D and E....make up cell theory

8 0

3 years ago

A car is going 8 meters per second on an access road into a highway

TiliK225 [7]

Answer:

20.96 m/s^2 (or 21)

Explanation:

Using the formula (final velocity - initial velocity)/time = acceleration, we can plug in values and manipulate the problem to give us the answer.

At first, we know a car is going 8 m/s, that is its initial velocity.

Then, we know the acceleration, which is 1.8 m/s/s

We also know the time, 7.2 second.

Plugging all of these values in shows us that we need to solve for final velocity. We can do so by manipulating the formula.

(final velocity - initial velocity) = time * acceleration

final velocity = time*acceleration + initial velocity

After plugging the found values in, we get 20.96 m/s/s, or 21 m/s

3 0

3 years ago

5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc

ludmilkaskok [199]

Answer:

Δθ₁ = 172.5 rev

Δθ₁h = 43.1 rev

Δθ₂ = 920 rev

Δθ₂h = 690 rev

Explanation:

Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:

$\Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2} (1)$

Now, we need first to find the value of the angular acceleration, that we can get from the following expression:

$\omega_{f1} = \omega_{o} + \alpha * \Delta t (2)$

Since the machine starts from rest, ω₀ = 0.
We know the value of ωf₁ (the operating speed) in rev/min.
Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:

$3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)$

Replacing by the givens in (2):

$57.5 rev/sec = 0 + \alpha * 6 s (4)$

Solving for α:

$\alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)$

Replacing (5) and Δt in (1), we get:

$\Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev (6)$

in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

$\Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev (7)$

In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

$\Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2} (8)$

First of all, we need to find the value of the angular acceleration during the second period.
We can use again (2) replacing by the givens:
ωf =0 (the machine finally comes to an stop)
ω₀ = ωf₁ = 57.5 rev/sec
Δt = 32 s

$0 = 57.5 rev/sec + \alpha * 32 s (9)$

Solving for α in (9), we get:

$\alpha_{2} =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)$

Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:

$\Delta \theta_{2} = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)$

In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
$\Delta \theta_{2h} = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)$

7 0

3 years ago

~~~!Here's the question!~~~

gtnhenbr [62]

The answer would most likely be A since obviously gravity weighs things down which helps the every other masses stay settled in place

6 0

3 years ago

Read 2 more answers