On which teeth are brackets most commonly

A block of mass = 4.00 kg is supported by a spring scale with unit of measure of Newtons. This spring scale is attached to top o

faust18 [17]

Answer:

118 N

Explanation:

Given mass of the block, m = 4.00kg.

The acceleration of the elevator, a = 3.0 m/s^2.

As elevotar attaced with spring scale and accelerating upward

(block and elevator), so total force

$F_N-mg=ma$

Here, mg is the weight of the block downward direction.

or

$F_N=ma+mg=m(g+a)$

substitute the given value, we get

$F_N=4kg(9.8m/s^2+3m/s^2)$

= 117.6 N = 118 N.

Thus, the reading on the spring scale to 3 significant figures is 118 N.

5 0

3 years ago

A school bus transporting students to school is an example of

jok3333 [9.3K]

Answer:

The answer is D. If I got this right will you please mark me as Brainliest.

3 0

3 years ago

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An astronaut spends 3 months in space and then returns to Earth. When she returns to Earth, her muscles have weakened. This is b

Pepsi [2]

False. there's less gravitational force in space than on earth

6 0

3 years ago

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A cylinder which is in a horizontal position contains an unknown noble gas at 4.63 × 104 Pa and is sealed with a massless piston

AleksandrR [38]

Answer:

The change in internal energy of the system is -17746.78 J

Explanation:

Given that,

Pressure $P=4.63\times10^{4}\ Pa$

Remove heat $\Delta U= -1.95\times10^{4}\ J$

Radius = 0.272 m

Distance d = 0.163 m

We need to calculate the internal energy

Using thermodynamics first equation

$dU=Q-W$ ...(I)

Where, dU = internal energy

Q = heat

W = work done

Put the value of W in equation (I)

$dU=Q-PdV$

Where, W = PdV

Put the value in the equation

$dU=-1.95\times10^{4}-(4.63\times10^{4}\times3.14\times(0.272)^2\times(-0.163))$

$dU=-17746.78\ J$

Hence, The change in internal energy of the system is -17746.78 J

3 0

3 years ago

The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second if i

Sergio [31]

Answer with Explanation:

We are given that

Diameter of fighter plane=2.3 m

Radius= $r=\frac{d}{2}=\frac{2.3}{2}=1.15 m$

a.We have to find the angular velocity in radians per second if it spins=1200 rev/min

Frequency= $\frac{1200}{60}=20 Hz$

1 minute=60 seconds

Angular velocity= $\omega=2\pi f$

Angular velocity= $2\times \frac{22}{7}\times 20=125.7 rad/s$

b.We have to find the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac.

$v=r\omega=1.15\times 125.7=144.56 m/s$

c.Centripetal acceleration= $\omega^2 r=(125.7)^2(1.15)=18170.56 m/s^2$

Centripetal acceleration== $\frac{18170.56\times g}{9.81}=1852.25 g m/s^2$

7 0

3 years ago