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3 years ago

John attached a ball to a spring

Physics

1 answer:

chubhunter [2.5K]3 years ago

8 0

Answer:

C down

Explanation:

The ball is pulling the spring down so this must be the direction of the force

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PLEASE ANSWER THIS ASAP!!!

Burka [1]

Answer:

I guess B

Explanation:

Because it stop by it self so that is digital

5 0

2 years ago

If a grocery cart with a mass of 16.5 kg accelerates at +2.31 m/s2 against a frictional force of -15.0 N, what is the applied fo

ziro4ka [17]

Call the applied force 'A'. (Clever ?)

The forces on the cart are 'A' forward and 15 N backward.

The net force on the art is (A-15) forward.

F = m a

Net forward force = (mass of the cart) x (its forward acceleration)

(A - 15) = (16.5) x (2.31)

A - 15 = 38.115

Add 15 to each side:

<u>A = 53.115 newtons</u>

5 0

3 years ago

Read 2 more answers

Sirius A has a luminosity of 26 LSun and a surface temperature of about 9400 K.What is its radius?

wlad13 [49]

Answer

Given,

Sirius A surface temperature,T = 9400 K

Sirius A luminosity,L = 26 L₀

L₀ is the luminosity of sun.

Radius of sun = 695700000 m

Temperature on sun surface = 5780 K

Luminous intensity is given by:-

$L=4 \pi R^{2} \sigma T^{4}$

Now

$\frac{L}{L_{0}}=\frac{4 \pi R^{2} \sigma T^{4}}{4 \pi R_{0}^{2} \sigma T_{0}^{4}}=26$

$\Rightarrow \frac{R^{2} T^{4}}{R_{0}^{2} T_{0}^{4}}=26$

$\Rightarrow R^{2}=26 \times \frac{R^{2} T_{0}^{4}}{T^{4}}=26 \times \frac{695700000^{2} \times 5780^{4}}{9400^{4}}$

$R=1341246640\ m=1.34 \times 10^{9}$

7 0

3 years ago

Suppose 8.41 moles of a monatomic ideal gas expand adiabatically, and its temperature decreases from 395 to 279 K. Determine (a)

sergeinik [125]

Answer:

a) W=12166.20876 J

b) U= -12166.20876 J

Explanation:

No. of moles, n = 8.41

Change of temperature, ΔT = T1 - T2

= 395 - 279

= 116 K

For monatomic gas, γ = 5/3

γ -1 = 2 /3

Solution:

(a)

Work done, $W= \frac{nR}{\gamma-1}(T_1-T_2)$

plugging values we get

$W= \frac{8.314\times8.41}{2/3}(116)$

Ans: 12166.20876 J

Work done, W = + 12166.20876 J

(b)

From first law of thermodynamics, dQ = U + W

but, dQ = 0 ( adiabatic process)

Hence, U = - W

= - 12166.20876 J

Ans:

Change in internal energy, U = - 12166.20876 J

8 0

3 years ago

A block is given a very brief push up a 20.0° frictionless incline to give it an initial speed of 12.0 m/s. (a) how far along th

shtirl [24]

The incline is frictionless, this means we can use the conservation of energy: the initial kinetic energy of the block
$K= \frac{1}{2}mv^2$
is converted into gravitational potential energy
$U=mgh$
where h is the height reached by the block as it stops. By equalizing the two formulas, we get
$\frac{1}{2} mv^2=mgh$
$h= \frac{v^2}{2g}= \frac{(12.0 m/s)^2}{2(9.81 m/s^2)} =7.3 m$

However, this is the maximum height reached by the block. The distance along the surface of the plane is given by:
$d= \frac{h}{\sin 20^{\circ}}= \frac{7.3 m}{\sin 20^{\circ}}=21.3 m$

4 0

4 years ago