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3 years ago

How do you do this question?

Physics

1 answer:

Arte-miy333 [17]3 years ago

5 0

Explanation:

The box has no acceleration in the perpendicular direction, so the net force in that direction is 0.

Sum of forces in the parallel direction:

∑F = f − F∥ − mg∥

∑F = f − F cos φ − mg sin θ

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How much extra water does a 147-lb concrete canoe displace compared to an ultralightweight 38-lb kevlar canoe of the same size c

romanna [79]

We use the formula, to calculate the volume of water displaced by concrete canoe,

$V =\frac{W}{\gamma }$

Here, W is the weight of concrete canoe and $\gamma$ is the specific weight of water and its value is $62.4\ lb/ft^3$ .

So,

$V =\frac{ 147\ lb}{62.4\ lb/ft^3}=2.356\ ft^3$ .

Now the volume of water occupied in ultra lightweight kevlar canoe,

$v=\frac{w}{\gamma}$

Here, w is weight of kevlar canoe.

So,

$v=\frac{38\ lb}{62.4\ lb/ft^3} =0.6089\ ft^3$

Thus, the volume of water displaced,

$=V-v=2.356\ ft^3-0.6089\ ft^3=2.19\ ft^3$ .

Hence, the volume of water displaced canoe compared to an ultra-lightweight kevlar canoe is $2.19\ ft^3$

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3 years ago

Hi please help with this question! Need the workings.

andreyandreev [35.5K]

<span>Mass of the copper penny m = 2.6 g Atomic mass of copper = 63.55, Atomic number = 29, So the number of neutrons = Atomic mass - Atomic number = 63 - 29 = 34 a. Neutron mass = 34 x (2.6 / 63.55) = 1.4 grams Copper atoms per mole = 6.040 x 10^23 atoms/mol moles of copper = 2.6 / 63.06 = 0.04123 mol Total atoms in the copper = 6.040 x 10^23 atoms/mol x 0.04123 mol = 0.25 x 10^23 atoms Number of electrons in the copper = 29 per atom Mass of the electron = 9.085 x 10^-28 g b. Electron mass = 0.25 x 10^23 x 29 x 9.085 x 10^-28 = 65.86 x 10^-5 g</span>

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3 years ago

A house is wired so that one electrical source comes to a room but many outlets and lights work from that source . when one ligh

Misha Larkins [42]

The room is wired as a parallel circuit

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4 years ago

Which statement explains why you are thrown forward when the car you are riding in suddenly stops ?

ozzi

Answer:

You are are in motion as the car is driving ñ, and when it stops, your body continues to be in motion

Explanation:

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3 years ago

Tutorial Exercise An unstable atomic nucleus of mass 1.83 10-26 kg initially at rest disintegrates into three particles. One of

kogti [31]

Answer:

A) v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) K_total = 373.08 × 10^(-15) J

Explanation:

We are given;

Mass of unstable atomic nucleus; M = 1.83 × 10^(-26) kg

Mass of first particle; m1 = 5.03 × 10^(-27) kg

Speed of first particle in y-direction; v1 = (6 × 10^(6) m/s) j^

Mass of second particle; m2 = 8.47 × 10^(-27) kg

Speed of second particle in x - direction; v2 = (4 × 10^(6) m/s) i^

Now, we don't have the mass of the third particle but since we are told the unstable atomic nucleus disintegrates into 3 particles, thus;

M = m1 + m2 + m3

1.83 × 10^(-26) = (5.03 × 10^(-27)) + (8.47 × 10^(-27)) + m3

m3 = (1.83 × 10^(-26)) - (13.5 × 10^(-27))

m3 = 4.8 × 10^(-27) kg

A) Applying law of conservation of momentum, we have;

MV = (m1 × v1) + (m2 × v2) + (m3 × v3)

Now, the unstable atomic nucleus was at rest before disintegration, thus V = 0 m/s.

Thus, we now have;

0 = (m1 × v1) + (m2 × v2) + (m3 × v3)

We want to find the velocity of the third particle v3. Let's make it the subject of the formula;

v3 = [(m1 × v1) + (m2 × v2)]/(-m3)

Plugging in the relevant values, we have;

v3 = [(5.03 × 10^(-27) × 6 × 10^(6))j^ + (8.47 × 10^(-27) × 4 × 10^(6))i^]/(-4.8 × 10^(-27))

v3 = [(30.18 × 10^(-21))j^ + (33.88 × 10^(-21))i^]/(-4.8 × 10^(-27))

v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) Formula for kinetic energy is;

K = ½mv²

Now,total kinetic energy is;

K_total = K1 + K2 + K3

K1 = ½ × 5.03 × 10^(-27) × (6 × 10^(6))²

K1 = 90.54 × 10^(-15) J

K2 = ½ × 8.47 × 10^(-27) × (4 × 10^(6))²

K2 = 67.76 × 10^(-15)

To find K3, let's first find the magnitude of v3 because it's still in vector form.

Thus;

v3 = √[(-6.29 × 10^(6))² + (-7.06 × 10^(6))²]

v3 = 9.46 × 10^(6) m/s

K3 = ½ × 4.8 × 10^(-27) × (9.46 × 10^(6))²

K3 = 214.78 × 10^(-15) J

K_total = (90.54 × 10^(-15)) + (67.76 × 10^(-15)) + (214.78 × 10^(-15))

K_total = 373.08 × 10^(-15) J

7 0

4 years ago