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4 years ago

If an item as a density of.63 g/mL, when placed in water the item will

Physics

1 answer:

sergij07 [2.7K]4 years ago

5 0

Float. This is because water has a density of 1 g/ml, so anything less than this will float.

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The oven draws 1100 W of power, and you have determined that it heats with an efficiency of 51%. Using this information, determi

LiRa [457]

Answer:

134.75935 seconds

Explanation:

m = Amount of water = 240 mL

c = Specific heat of water = 4.2 J/mL °C

$\Delta T$ = Change in temperature = $100-25=75$

t = Time taken

P = Power = 1100 W

E = Pt

Efficiency = 51 %

So, Energy = 0.51Pt

As energy is conserved

Q = 0

$\\\Rightarrow -mc\Delta T+E=0\\\Rightarrow -mc\Delta T+Pt=0\\\Rightarrow -240\times 4.2(100-25)+0.51\times 1100t=0\\\Rightarrow t=\frac{240\times 4.2(100-25)}{0.51\times 1100}\\\Rightarrow t=134.75935\ s$

The it takes to raise the temperature is 134.75935 seconds

6 0

4 years ago

¿Qué trabajo hace una fuerza de 110 N cuando mueve su punto de aplicación 20 mt en su misma dirección? *

lys-0071 [83]

Answer:

W = 2.200 Joules

Explanation:

Datos (data):

Fuerza [force] (F) = 110 N
Metros [meters] (m) = 20 m
Trabajo [work] (W) = ?

Usar la fórmula (use formula):

$\boxed{\bold{W = F*d}}$

Reemplazar (replace):

$\boxed{\bold{W = 110\ N*20\ m}}$

Resolver la multiplicación, recuerda que 1 N * 1 m = 1 J (resolve the multiplication, remember that 1 N * 1 m = 1 J:

$\boxed{\boxed{\bold{W =2.200\ J}}}$

Greetings.

7 0

3 years ago

Using hookes law find the distance a spring with am elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it

siniylev [52]

484848484 48v88 848 48gv

6 0

3 years ago

1. An electron (Q=16x10^-20 C, m=1x10^-30 kg) moving at half a megameter per second up the page enters a region with a uniform m

jek_recluse [69]

Explanation:

It is given that,

Charge on electron, $q=16\times 10^{-20}\ C$

Mass of the electron, $m=9.1\times 10^{-31}\ kg$

Speed of the electron, $v=0.5\ Mm/s=0.5\times 10^6\ m/s$

Magnetic field, B = 1 T (directed out of the page)

Let F is the magnetic force acting on the electron. It is given by :

$F=qvB\ sin\theta$

Here, $\theta=90^{\circ}$

$F=qvB$

$F=16\times 10^{-20}\ C\times 0.5\times 10^6\ m/s\times 1\ T$

$F=8\times 10^{-14}\ N$

Using the right hand rule, the direction of magnetic force is upward to the plane of the paper. Also, the electron will follow the circular path. It is given by :

$r=\dfrac{mv}{qB}$