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3 years ago

13

When the balloon sticks to the wall (assuming it sticks to the wall). It is

1 answer:

inn [45]3 years ago

8 0

The answer is false

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A 2.0-kg block sliding on a rough horizontal surface is attached to one end of a horizontal spring (k = 250 N/m) which has its o

Burka [1]

Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.

There are two forces performing work on the block: the restoring force of the spring and kinetic friction.

By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:

$W_{\rm total}=\Delta K$

or

$W_{\rm friction}+W_{\rm spring}=0-K=-K$

where <em>K</em> is the block's kinetic energy at the equilibrium point,

$K=\dfrac12\left(2.0\,\mathrm{kg}\right)\left(2.6\dfrac{\rm m}{\rm s}\right)^2=6.76\,\mathrm J$

Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is

$W_{\rm spring}=-\dfrac12\left(250\dfrac{\rm N}{\rm m}\right)(0.20\,\mathrm m)^2=-5.00\,\mathrm J$

Compute the work performed by friction:

$W_{\rm friction}-5.00\,\mathrm J=-6.76\,\mathrm J \implies W_{\rm friction}=-1.76\,\mathrm J$

By Newton's second law, the net vertical force on the block is

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0 ==> <em>n</em> = <em>mg</em>

where <em>n</em> is the magnitude of the normal force from the surface pushing up on the block. Then if <em>f</em> is the magnitude of kinetic friction, we have <em>f</em> = <em>µmg</em>, where <em>µ</em> is the coefficient of kinetic friction.

So we have

$W_{\rm friction}=-f(0.20\,\mathrm m)$

$\implies -1.76\,\mathrm J=-\mu\left(2.0\,\mathrm{kg}\right)\left(9.8\dfrac{\rm m}{\mathrm s^2}\right)(0.20\,\mathrm m)$

$\implies \boxed{\mu\approx0.45}$

4 0

3 years ago

It turns out that Mercury and Mars have the same gravity as one another – that is, you would weigh the same on the surface of Me

Tamiku [17]

Answer:

Explanation:

The value of acceleration due to gravity of mars is same as that the value of acceleration due to gravity of Mercury.

The value of acceleration due to gravity of a planet depends on its mass and the radius.

The formula for the acceleration due to gravity is given by

$g=\frac{GM}{R^{2}}$

Where, M be the mass of the planet, R be the radius of the planet.

As according to the question, the value of acceleration due to gravity for Mercury is same as that of Mars.

$\frac{GM_{mercury}}{R_{mercury}^{2}}=\frac{GM_{mars}}{R_{mars}^{2}}$

$\frac{M_{mercury}}{R_{mercury}^{2}}=\frac{M_{mars}}{R_{mars}^{2}}$

As teh radius of Mercury is small, and acceleration due to gravity is same, it is possible because the mass of Mercury is more than the mass of Mars.

4 0

3 years ago

Austin invested $11,000 in an account paying an interest rate of 5.7% compounded quarterly. Assuming no deposits or withdrawals

daser333 [38]

Answer:

15448

Explanation:

Compounded Quarterly:

A=P\left(1+\frac{r}{n}\right)^{nt}

A=P(1+

n

r

)

nt

Compound interest formula

P=11000\hspace{35px}r=0.057\hspace{35px}t=6\hspace{35px}n=4

P=11000r=0.057t=6n=4

Given values

A=11000\left(1+\frac{0.057}{4}\right)^{4(6)}

A=11000(1+

4

0.057

)

4(6)

Plug in values

A=11000(1.01425)^{24}

A=11000(1.01425)

24

Simplify

A=15448.0290759

A=15448.0290759

Use calculator

3 0

3 years ago

Read 2 more answers

A small 16 kilogram canoe is floating downriver at a speed of 4 m/s. What is the canoe's kinetic energy?

seraphim [82]

Kinetic Energy,K.E=1/2MV²

mass,m=16kg

velocity,v=4m/s

K.E=1/2×16×4²

=128kgm²/s²

=128 Joules

5 0

3 years ago

Read 2 more answers

You throw a 50.0g blob of clay directly at the wall with an initial velocity of -5.00 m/s i. The clay sticks to the wall, and th

Whitepunk [10]

Answer:0.25 kg-m/s

Explanation:

Given

mass of blob $m=50 gm$

initial velocity $u=-5 m/s\ \hat{i}$

time of collision $t=20 ms$

we know Impulse is equal to change in momentum

initial momentum $P_i=mu$

$P_i=50\times 10^{-3}\times (-5)=-0.25 kg-m/s$

Final momentum $P_f=50\times 10^{-3}v$

$P_f=0$ as final velocity is zero

Impulse $J=P_f-P_i$

$J=0-(-0.25)$

$J=0.25 kg-m/s$

5 0

3 years ago