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3 years ago

10

A car driving at 25 m/s, travels 318 m. How long does it take?

1 answer:

labwork [276]3 years ago

7 0

Answer:

12.72 sec

Explanation:

The time it takes to travel a certain distance can be found by dividing the distance that needs to be travelled by the speed with which you are travelling, therefore...

$Time = \frac{Distance}{Speed} = \frac{318}{25} = 12.72 sec$

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What force, in newtons, must you exert on the balloon with your hands to create a gauge pressure of 62.5 cm H2O, if you squeeze

yulyashka [42]

Answer:54.70 N

Explanation:

Given

Gauge Pressure of $62.5 cm of H_2O$

i.e. $h=62.5 cm =0.625 m$

Effective area $A=51 cm^2$

initial Pressure $= 1 atm=101.325 kPa$

Gauge Pressure $P=\rho gh$

$\rho =density\ of\ water =1000 kg/m^3$

$P_{gauge}=1000\times 9.8\times 0.625=5.937 kPa$

Force creates a pressure of $P_1$ which will be equal to Gauge Pressure

$P_1=\frac{F}{A}$

$P_1=P_{gauge}$

$\frac{F}{A}=5.937 kPa$

$F=5.937\times 51\times 10^{-4}\times 10^3$

$F=30.27 N$

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4 years ago

Which piece of evidence did Alfred Wegener use to develop the theory of continental drift?

Eduardwww [97]

The right answer for the question that is being asked and shown above is that: "A.tectonic activity concentrated in certain areas." A piece of evidence did Alfred Wegener use to develop the theory of continental drift is that <span>A.tectonic activity concentrated in certain areas</span>

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3 years ago

astra-53 [7]

Answer:

1968

Explanation:

2400*20.5*0.004

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3 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

4 years ago

A 250 kg cart is at the top of a hill that is 32 m high, what is its potential energy?

atroni [7]

Answer:

<h2>80,000 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 250 × 10 × 32

We have the final answer as

<h3>80,000 J</h3>

Hope this helps you

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3 years ago