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4 years ago

A ball is thrown at a 60.0° angle above the horizontal across level ground. It is released from a

1 answer:

6 0

Write out what you have which is:
initial velocity
final velocity
Y distance
degree

You do not have :
a
X distance
t

from what you have you can plug into your formulas to get time.

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An ideal spring is hung vertically from the ceiling. The spring constant is k = 125 N/m. A block of mass m = 650 g (1000 g = 1 k

Brrunno [24]

Answer:

0.102 m

Explanation:

k = spring constant of the spring = 125 N/m

m = mass of the block attached to the spring = 650 g = 0.650 kg

x = maximum extension of the spring

h = height dropped by the block = x

Using conservation of energy

Spring potential energy gained = Gravitational potential energy lost

(0.5) k x² = mgh

(0.5) k x² = mgx

(0.5) (125) x = (0.650) (9.8)

x = 0.102 m

3 0

3 years ago

En el MCU, la aceleración es: a) constante b) Varía en módulo, direccion, y sentido c) constante solo en el módulo d) constante

JulijaS [17]

If you put it into English in the comments I would be more then happy to help you! Thank you!

7 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

4 years ago

An open system starts with 52 J of mechanical energy. The energy changes

spayn [35]

Answer:

I think the answer is D,54 joules

5 0

3 years ago

Need amswer fast plz

FrozenT [24]

Answer:

I need this for may schooling

6 0

2 years ago