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2 years ago

What is the weight of a 1-kilogram brick resting on a table?

Physics

1 answer:

MakcuM [25]2 years ago

5 0

Answer:

The weight if the block is 10Newtons

Explanation:

The weight of any object is quantity of matter the object contains and it is always acting downwards on such body. This shows that the object is under the influence of gravity.

The weight of an object is calculated as mass of the object × its acceleration due to gravity

W = mg

Give the mass of the brick to be 1kg

g is the acceleration due to gravity = 10m/s²

Weight of the object = 1 × 10

= 10kgm/s² or 10Newtons

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Find the lowest two frequencies that produce a maximum sound intensity at the positions of Moe and Curly.

Mars2501 [29]

Answer:

hello your question has some missing parts below is the complete question

and the missing diagram

The two speakers emit sound that is 180° out of phase and of a single frequency,ƒ, Find the lowest two frequencies that produce a maximum sound intensity at the positions of Moe and Curly.

answer : 1316.2 hertz

Explanation:

The frequency that produce the maximum sound intensity can be calculated using the relation below

dsin ∅ = n A

where A = dsin ∅ / n when n = 1 . d = 0.800

A = 0.800 * ( 1 / 3.162 )

A = 0.253 m

speed of sound = 333 m/s

frequency = speed / A

= 333 / 0.253 = 1316.2 hertz

7 0

2 years ago

A 23.3-kg mass is attached to one end of a horizontal spring, with the other end of the spring fixed to a wall. the mass is pull

White raven [17]

For a simple harmonic motion, the position of the mass at any time t is given by
$x(t)=A cos (\omega t)$
where
A is the amplitude of the motion (in this problem, A=17.5 cm)
$\omega$ is the angular frequency of the oscillator
t is the time

The angular frequency of the motion in the problem is given by
$\omega = 2 \pi f= 2 \pi (8.38 Hz) = 52.7 rad/s$

And so, we can find the position x of the mass (with respect to the equilibrium position) at time t=2.50 s:
$x(2.50 s)=(17.5 cm) \cos ( (52.7 rad/s)(2.50 s))=17.2 cm$

6 0

2 years ago

Read 2 more answers

Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 m3>s. (a) How fast will it shoot out of a hole 4

kati45 [8]

Answer:

velocity = 472 m/s

velocity = 52.4 m/s

Explanation:

given data

steady rate = 0.750 m³/s

diameter = 4.50 cm

solution

we use here flow rate formula that is

flow rate = Area × velocity .............1

0.750 = $\frac{\pi }{4}$ × (4.50× $10^{-2}$ )² × velocity

solve it we get

velocity = 472 m/s

and

when it 3 time diameter

put valuer in equation 1

0.750 = $\frac{\pi }{4}$ × 3 × (4.50× $10^{-2}$ )² × velocity

velocity = 52.4 m/s

5 0

2 years ago

You are on vacation in San Francisco and decide to take a cable car to see the city. A 5800-kgkg cable car goes 260 mm up a hill

Stella [2.4K]

Answer:

$4.325\times10^6J$

Explanation:

Mass of the cable car, m = 5800 kg

It goes 260 m up a hill, along a slope of $\theta=17^o$

Therefore vertical elevation of the car = $260sin\theta=260sin17^o=76.0166m$

Now, when you get into the cable car, it's velocity is zero, that is, initial kinetic energy is zero (since K.E. = $\frac{1}{2} mv^2$ ). Similarly as the car reaches the top, it halts and hence final kinetic energy is zero.

Therefore the only possible change in the cable car system is the change in it's gravitational potential energy.

Hence, total change in energy = mgh = $5800\times9.81\times76.0166J=4.325\times10^6J$

where, g = acceleration due to gravity

h = height/vertical elevation

4 0

2 years ago

Inna Hurry is traveling at 6.8 m/s, when she realizes she is late for an appointment. She accelerates at 4.5 m/s^2 for 3.2 s. Wh

Alborosie

Answer:

1) v = 21.2 m/s

2) S = 63.33 m

3) s = 61.257 m

4) Deceleration, a = -4.32 m/s²

Explanation:

1) Given,

The initial velocity of Inna, u = 6.8 m/s

The acceleration of Inna, a = 4.5 m/s²

The time of travel, t = 3.2 s

Using the first equation of motion, the final velocity is

v = u + at

= 6.8 + 4.5 x 3.2

= 21.2 m/s

The final velocity of Inna is, v = 21.2 m/s

2) Given,

The initial velocity of Lisa, u = 12 m/s

The final velocity of Lisa, v = 26 m/s

The acceleration of Lisa, a = 4.2 m/s²

Using the III equations of motion, the displacement is

v² = u² +2aS

S = (v² - u²) / 2a

= (26² -12²) / 2 x 4.2

= 63.33 m

The distance Lisa traveled, S = 63.33 m

3) Given,

The initial velocity of Ed, u = 38.2 m/s

The deceleration of Ed, d = - 8.6 m/s²

The time of travel, t = 2.1 s

Using the II equations of motion, the displacement is

s = ut + 1/2 at²

=38.2 x 2.1 + 0.5 x(-8.6) x 2.1²

= 61.257 m

Therefore, the distance traveled by Ed, s = 61.257 m

4) Given,

The initial velocity of the car, u = 24.2 m/s

The final velocity of the car, v = 11.9 m/s

The time taken by the car is, t = 2.85 s

Using the first equations of motion,

v = u + at

∴ a = (v - u) / t

= (11.9 - 24.2) / 2.85

= -4.32 m/s²

Hence, the deceleration of the car, a = = -4.32 m/s²

5 0

2 years ago

Read 2 more answers