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3 years ago

If an object starts out at rest and accelerates to 100 m/s, what is its initial speed?

1 answer:

6 0

No. Sadly, you're not right. The whole problem here is the meaning of a few words. "Starts out"is the same thing as "initial" conditions. "At rest" is the same thing as not moving ... zero speed. So if an object starts out at rest, that says that its initial speed is zero ... choice 'b'.

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Water falls without splashing at a rate of 0.200 L/s from a height of 3.60 m into a 0.730 kg bucket on a scale. If the bucket is

dimaraw [331]

Answer:

15.106 N

Explanation:

From the given information,

The weight of the bucket can be calculated as:

$W_b = m_bg = \\ \\ W_b = (0.730 \ kg) ( 9.80 \ m/s^2) \\ \\ W_b = 7.154 \ N$

The mass of the water accumulated in the bucket after 3.20s is:

$m_w= (0.20 \ L/s) ( 3.20)s$

$m _w=0.64 \ kg$

To determine the weight of the water accumulated in the bucket, we have:

$W_w = m_w g$

$W_w = ( 0.64 \ kg )(9.80\ m \ /s^2)$

$W_w = 6.272 \ N$

For the speed of the water before hitting the bucket; we have:

$v = \sqrt{2gh}$

$v = \sqrt{2*9.80 \ m/s^2 * 3.60 \ m}$

v = 8.4 m/s

Now, the force required to stop the water later when it already hit the bucket is:

$F = v ( \dfrac {dm}{dt} )$

$F = (8.4 \ m/s)( 0.200 \ L/s)$

F = 1.68 N

Finally, the reading scale is:

$F_{scale$ = 7.154 N + 6.272 N + 1.68 N

= 15.106 N

6 0

2 years ago

Read 2 more answers

what is the main difference between a substance going through a physical change and one going through a chemical ?

Ainat [17]

Answer: Physical changes only change the appearance of a substance, not its chemical composition. Chemical changes cause a substance to change into an entirely substance with a new chemical formula. Chemical changes are also known as chemical reactions.

Explanation:

3 0

2 years ago

Read 2 more answers

A ___ is the unit of measurement for force.

bogdanovich [222]

Answer:

Newton (N)

Explanation:

A newton is the unit of measurement for force

4 0

3 years ago

A charge q of magnitude 6.4 × 10^-19 coulombs moves from point A to point B in an electric field of 6.5 × 10^4 newtons/coulomb.

lana [24]

In this problem we have the electric field intensity E:

E = 6.5 × $10^4$ newtons/coulomb

We have the magnitude of the load:

q = 6.4 × $10 ^{-19}$ coulombs

We also have the distance d that the load moved in a direction parallel to the field 1.2 × $10^{-2}$ meters.

We know that the electric potential energy (PE) is:

PE = qEd

So:

PE = (6.4 × $10^{-19}$ )(6.5 × $10^4$ )(1.2 × $10^{-2}$ )

PE = 5.0 x $10^{-16}$ joules

None of the options shown is correct.

6 0

3 years ago

A simple pendulum has length of 820mm. Calculate the frequency (g = 9.8 ms -2)

Vadim26 [7]

Answer:

$\huge\boxed{\sf f=0.55 \ Hz}$

Explanation:

Given Data:

Length = l = 820 mm = 0.82 m

Acceleration due to gravity = g = 9.8 ms⁻²

Required:

Frequency = f = ?

Formula:

$\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} }$

Solution:

$\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} } \\\\Put\ the\ givens\\\\f=\frac{1}{2 \pi} \sqrt{\frac{9.8}{0.82} }\\\\ f = 0.159 \times \sqrt{11.95} \\\\f=0.159 \times 3.457\\\\f=0.55 \ Hz\\\\\rule[225]{225}{2}$

7 0

2 years ago