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IRISSAK [1]
3 years ago
6

How many significant figures does 0.09164500561 have?

Physics
1 answer:
kolbaska11 [484]3 years ago
7 0

Answer:

10 Sig Figs

Explanation:

Just start counting at the first non zero after the decimal so in this case the nine, and count all of the numbers including zeros after that.

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A body is projected vertically upwards with a speed 100 m/s from the ground then distance it covers in its last second of its up
Valentin [98]

The distance travelled by the body on last second of its upwards journey is 4.935m

What is projectile motion?

A projectile is an object or particle that is launched toward the surface of the Earth and moves along a curved path only under the influence of gravity.

To solve this question, consider the horizontal distance to be H and

distance traveled by the body in the last second be d

H=U^{2} /2g

H=100^{2} /2*9.8

H=510.2m

Now time taken to reach a height of

510.2 m= \sqrt[n]{2H/g}

=10.2 seconds

Now,

s=ut+1/2gt^{2}

H-d= (100)(t-1)-1/2g(t-1)^{2}

510.2-d=100*9.2-1/2*9.8*9.2^{2}

d=4936m

So, the correct answer is '4.936m'.

To know more about projectile motion visit:

brainly.com/question/11049671

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8 0
2 years ago
a 0.05-kg starts from rest at a height of 0.95m. assuming no friction, what is the kinetic energy of the car when it reaches the
Alex73 [517]
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3 0
3 years ago
A 12 oz can of soda is left in a car on a hot day. In the morning the soda temperature was 60oF with a gauge pressure of 40 psi.
Neko [114]
In this case, volume of the can remains constant. The relationship between pressure and temperature at constant volume is given by:

P/T = Constant

Then
\frac{ P_{1} }{ T_{1} } = \frac{ P_{2} }{ T_{2} }

Where
P1 = 40 psi
P2 = ?
T1 = 60°F ≈ 289 K
T2 = 90°F ≈ 305 K (note, 363 K is not right)

Substituting;
P_{2} = \frac{ P_{1}  T_{2} }{ T_{1} } = \frac{40*305}{289}  =42.21 psi
3 0
3 years ago
Steam is to be condensed on the shell side of a heat exchanger at 150 oF. Cooling water enters the tubes at 60 oF at a rate of 4
zalisa [80]

Answer:

a. 572Btu/s

b.0.1483Btu/s.R

Explanation:

a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.

From table A-3E, the specific heat of water is c_p=1.0\ Btu/lbm.F, and the steam properties as, A-4E:

h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R

Using the energy balance for the system:

\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s

b. Heat gained by the water is equal to the heat lost by the condensing steam.

-The rate of steam condensation is expressed as:

\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

4 0
3 years ago
Why do plane mirrors and convex mirrors form only virtual images?
Alekssandra [29.7K]
<span>B. because they cannot make light rays converge</span>
7 0
3 years ago
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