In a molecule of water, how many electrons are shared between each hydrogen atom and an oxygen atom? one two four

A 1500 kg car traveling east at 40 km/hr turns a corner and speeds up to a velocity of 50 km/hr due north. What is the change in

pantera1 [17]

Answer:

96046 Ns.

Explanation:

We shall represent velocity in vector form considering east direction as + ve x axis and north as + y direction.

40 km/h in the east

V₁ = 40 i

V₂ = 50j

momentum p₁ = mV₁

= 1500 X 40 i

= 60000 i

Momentum p₂ = mV₂

= 1500 X 50j

= 75000 j

Change in momentum

p₂ - p₁

75000j - 60000i

Magnitude of change

= $\sqrt{(750000)^2 +(60000)^2$

= 96046 Ns.

5 0

3 years ago

A backyard swimming pool with a circular base of diameter 6.00m is filled to depth 1.50m. (b) Two persons with combined mass 150

Mashutka [201]

The pressure increase at the bottom of the pool after they enter the pool and float is 106.103 Pa.

<h3>What is absolute pressure?</h3>

Absolute pressure is the force that exists in a space when there is no matter present, or when there is a perfect vacuum. This absolute zero serves as the baseline for measurements in absolute pressure. The measurement of barometric pressure is the greatest illustration of an absolute referenced pressure. In order to determine absolute pressure, a complete vacuum is used. In contrast, gauge pressure is the amount of pressure that is measured in relation to atmospheric pressure, also referred to as barometric pressure.

given,

diameter = 6 m

depth = h = 1.5 m

Atmospheric pressure = P₀ = 10⁵ Pa

a) absolute pressure

P = P₀ + ρ g h

P = 10⁵ + 1000 x 10 x 1.5

P = 1.15 x 10⁵ Pa

b) When two person enters into the pool,

mass of the two person = 150 Kg

weight of water level displaced exists equal to the weight of person.

$\rho \mathrm{Vg}=2 \mathrm{mg} \\$

$V=\frac{2 m}{\rho} \\$

$V=\frac{2 \times 150}{1000} \\$

$\mathrm{~V}=0.3 \mathrm{~m}^3$

Area of pool $=\frac{\pi}{4} d^2$

$&=\frac{\pi}{4} \times 6^2 \\$

$&=28.27 \mathrm{~m}^2$

Height of the water rise

$h &=\frac{V}{A} \\$

$h &=\frac{0.3}{28.27} \\$

$& \mathrm{~h}=0.0106 \mathrm{~m}$

Pressure increased

P = ρ g h

P = 1000 x 10 x 0.0106

P = 106.103 Pa

To learn more about absolute pressure refer to:

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4 0

1 year ago

An ideal ammeter would have zero resistance while an ideal voltmeter would have infinite resistance, why?.

Dafna1 [17]

Answer: An ideal ammeter would have zero resistance, because to ensure that, there is no voltage drop due to the internal resistance. Similarly, an ideal voltmeter would have infinite resistance, because to ensure that there is no current is drawn by the voltmeter.

Explanation: To find the answer, we need to know about the Ammeter and Voltmeter.

<h3>What is an ammeter?</h3>

An ammeter is a device, that can be used to measure the electric current flows through a circuit in amperes.

An ideal ammeter would have zero resistance, because to ensure that, there is no voltage drop due to the internal resistance when it is connected in series to measure the current.

<h3>What is voltmeter?</h3>

A voltmeter is a device, that can be used to measure the electric potential difference generated between the terminals of an electric circuit in volts.

An ideal voltmeter would have infinite resistance, because to ensure that there is no current is drawn by the voltmeter, when it is connected in parallel to measure the voltage.

Thus, we can conclude that, an ideal ammeter would have zero resistance, because to ensure that, there is no voltage drop due to the internal resistance. Similarly, an ideal voltmeter would have infinite resistance, because to ensure that there is no current is drawn by the voltmeter.

Learn more about the ammeter and voltmeter here:

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6 0

1 year ago

Adam is teeing off on hole number two. The hole is 390 yards away. It is a par four hole. What club should he use to tee off? Ex

kkurt [141]

Answer:

A driver.

Explanation:

Using a driver while at least 350 yds away is better than using a iron, because it will be a waste of the par 4 as it is not as powerful as the driver.

3 0

3 years ago

A stationary police car emits a sound of frequency 1240 HzHz that bounces off of a car on the highway and returns with a frequen

Tju [1.3M]

Answer

given,

frequency from Police car= 1240 Hz

frequency of sound after return = 1275 Hz

Calculating the speed of the car = ?

Using Doppler's effect formula

Frequency received by the other car

$f_1 = \dfrac{f_0(u + v)}{u}$ ..........(1)

u is the speed of sound = 340 m/s

v is the speed of the car

Frequency of the police car received

$f_2= \dfrac{f_1(u)}{u-v}$

now, inserting the value of equation (1)

$f_2= f_0\dfrac{u+v}{u-v}$

$1275=1240\times \dfrac{340+v}{340-v}$

1.02822(340 - v) = 340 + v

2.02822 v = 340 x 0.028822

2.02822 v = 9.799

v = 4.83 m/s

hence, the speed of the car is equal to v = 4.83 m/s

5 0

3 years ago