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3 years ago

A 700 kg racecar slowed from 30 m/s to 15 m/s. What is the change in momentum

1 answer:

8 0

Momentum = (mass) x (speed)

Original momentum = (700 kg) x (30 m/s) = 21,000 kg-m/s

Final momentum = (700 kg) x (15 m/s) = 10,500 kg-m/s

Change in momentum = - 10,500 kg-m/s .

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Anyone know the answer to this, would kindly appreciate it

Ipatiy [6.2K]

Answer:

The found value of r is:

r = 94.7

Explanation:

We know that

cosθ = Base / Hypotenuse

θ is the always the angle between base and hypotenuse

In this right angles triangle, the angle θ is given 63°.

Which means that the Base is 43 and the hypotenuse is r

Base = 43

Hypotenuse = r

Substitute θ=63°, Base = 43 and hypotenuse = r in the formula of cosθ mentioned above:

cos 63° = 43/r

Rearrange

r = 43/cos63°

r = 43/0.454

r= 94.7

6 0

3 years ago

Two lamps rated 60W; 240V and 100W, 240Vrespectively are connected in series to a 240V power source. Calculate;

nikdorinn [45]

Answer:

See the answers below.

Explanation:

The total power of the circuit is equal to the sum of the powers of each lamp.

$P=60+100\\P=160 [W]$

Now we have a voltage source equal to 240 [V], so by means of the following equation we can find the current circulating in the circuit.

$P=V*I$

where:

P = power [W]

V = voltage [V]

I = current [amp]

$I = P/V\\I=160/240\\I=0.67 [amp]$

So this is the answer for c) I = 0.67 [amp]

We know that the voltage of each lamp is 240 [V]. Therefore using ohm's law which is equal to the product of resistance by current we can find the voltage of each lamp.

$V=I*R$

where:

V = voltage [V]

I = current [amp]

R = resistance [ohms]

Therefore we replace this equation in the first to have the current as a function of the resistance and not the voltage.

$P=V*I\\and\\V = I*R\\P = (I*R)*I\\P=I^{2}*R$

$60 = (0.67)^{2}*R\\R_{60}=133.66[ohm] \\and\\100=(0.67)^{2} *R\\R_{100}=100/(0.66^{2} )\\R_{100}=225 [ohm]$

The effective resistance of a series circuit is equal to the sum of the resistors connected in series.

$R = 133.66 + 225\\R = 358.67 [ohms]$

7 0

3 years ago

What is the concentration of H+ ions at a pH = 8?

skelet666 [1.2K]

Taking into account the definition of pH and pOH, the concentration of H⁺ ions at pH=8 is 1×10⁻⁸ M and the concentration of OH⁻ ions at pH=8 is 1×10⁻⁶ M.

<h3>Definition of pH</h3>

pH is a measure of acidity or alkalinity that indicates the amount of hydrogen ions present in a solution or substance.

The pH is defined as the negative base 10 logarithm of the activity of hydrogen ions, that is, the concentration of hydrogen ions or H₃O⁺:

pH= - log [H⁺]= - log [H₃O⁺]

<h3>Definition of pOH</h3>

Similarly, pOH is a measure of hydroxyl ions in a solution and is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:

pOH= - log [OH⁻]

<h3 /><h3>Relationship between pH and pOH</h3>

The following relationship can be established between pH and pOH:

pOH + pH= 14

<h3>Concentration of H⁺ ions</h3>

In this case, pH=8. Replacing in the definition of pH:

8= - log [H⁺]

Solving:

Finally, the concentration of H⁺ ions at pH=8 is 1×10⁻⁸ M.

<h3>Concentration of OH⁻ ions</h3>

Being pH= 8, pOH is calculated as:

pOH + 8= 14

pOH= 14 - 8

pOH= 6

Replacing in the definition of pOH the concentration of OH⁻ ions is obtained:

- log [OH⁻]= 6

Solving

Finally, the concentration of OH⁻ ions at pH=8 is 1×10⁻⁶ M.

Learn more about pH and pOH:

brainly.com/question/16032912

brainly.com/question/13557815

#SPJ1

7 0

2 years ago

How to find the magnitude and direction of an accelerating object?

Dahasolnce [82]

I'm pretty sure you can find it out by using a speed monitor and compass or just observing it

if there any answer choices tell me.

<h2>HOPE I HELPED!!!</h2>
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4 0

3 years ago

A satellite orbits the Earth in an elliptical orbit. At perigee its distance from the center of the Earth is 22500 km and its sp

aleksley [76]

Answer:

$6.09294\times 10^{24}\ kg$

Explanation:

K = Kinetic energy

$v_p$ = Perigee speed = 4280 m/s

$v_a$ = Apogee speed = 3990 m/s

$r_p$ = Perigee Distance = 22500000 m

$r_a$ = Apogee Distance = 24100000 m

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Earth

m = Mass of satellite

In this system the kinetic and potential energies are conserved

$K_p+P_p=K_a+P_a\\\Rightarrow \frac{1}{2}mv_p^2-\frac{GMm}{r_p}=\frac{1}{2}mv_a^2-\frac{GMm}{r_a}\\\Rightarrow \frac{1}{2}m(v_p^2-v_a^2)+GMm\left(\frac{1}{r_a}-\frac{1}{r_p}\right)=0\\\Rightarrow M=\frac{v_a^2-v_p^2}{2G}\times \left(\frac{1}{r_a}-\frac{1}{r_p}\right)^{-1}\\\Rightarrow M=\frac{3990^2-4280^2}{2\times 6.67\times 10^{-11}}\times \left(\frac{1}{24100000}-\frac{1}{22500000}\right)^{-1}\\\Rightarrow M=6.09294\times 10^{24}\ kg$

The mass of the Earth is $6.09294\times 10^{24}\ kg$

3 0

4 years ago