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Karolina [17]
2 years ago
8

Find the slope of the line containing the two points (1,−1) and (−5,−3). (1 point)

Physics
1 answer:
grandymaker [24]2 years ago
4 0

Answer:

(-4,-2)

Explanation:

You might be interested in
A particle with mass 1.81*10^-3kg and a charge of 1.22*10^-8C has, at a given instant, a velocity v= (3.0*10^4 m/s)j.
charle [14.2K]

Answer:

a = -0.33 m/s² k^

Direction: negative

Explanation:

From Newton's law of motion, we know that;

F = ma

Now, from magnetic fields, we know that;. F = qVB

Thus;

ma = qVB

Where;

m is mass

a is acceleration

q is charge

V is velocity

B is magnetic field

We are given;

m = 1.81 × 10^(−3) kg

q = 1.22 × 10 ^(−8) C

V = (3.00 × 10⁴ m/s) ȷ^.

B = (1.63T) ı^ + (0.980T) ȷ^

Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;

a = qVB/m

a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))

From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^

Thus;

a = -0.33 m/s² k^

7 0
2 years ago
When an oxygen atom forms an ion, it gains two electrons. What is the electrical charge of the oxygen ion?
Mashutka [201]
-2. This is because it is gaining electrons. Electrons are negative. Therefore it is a negative charge.
6 0
3 years ago
What is the kinetic energy of a vehicle that has a mass of 3,500 kg and is moving at 40 m/s
stiv31 [10]

Answer:2800000j

Explanation:

For us to know the kinetic energy of the vehicle,

Where m is the mass

And v is the velocity

Then, K.E=1/2mv^2

While, K.E=1/2×3500×40^2

Therefore, our answer will now be

K.E=2800000j

5 0
3 years ago
The nose of an ultralight plane is pointed south, and its airspeed indicator shows 28 m/s . the plane is in a 18 m/s wind blowin
leva [86]
<span>Here I think you have to find the velocity in x and y components where x is east and y is north
 So as air speed indicator shows the negative speed in y component and adding it in
  air speed while multiplying with the direction component we will get the velocity as velocity is a vector quantity so direction is also required
 v=-28 m/s y + 18 m/s (- x/sqrt(2) - y/sqrt(2))
 solving
  v= -12.7 m/s x-40.7 m/s y
 if magnitude of velocity or speed is required then
  speed= sqrt(12.7^2 + 40.7^2)
 speed= 42.63 m/s
 if angle is asked
  angle = arctan (40.7/12.7)
 angle = 72.67 degrees south of west</span>
6 0
3 years ago
If all this energy is added to 50 kg of water (the amount of water in a 165-lb person) at 37 ∘C, what is the final state of the
Reika [66]

Answer:

Vapors

Explanation:

We take into account that all the energy from the lightning has been transformed into steam.

\Delta U = Q -  W\\Q = mC \Delta T\\Q = mL

We calculate the amount of energy required by water to convert into steam.

Q_{water} = 50 \times \times 4180 \times (100-37)\\= 1.132 \times 10^7 \ J

Q_{change\ to\ steam} = 50 \times 2.256 \times 106 \\= 1.128 \times 10^8 \ J

Q_{total} = 1.132 \times 10^7 + 1.128 \times 10^8\\= 1.126 \times 10^8 \ J

From the lightning we received 10^{10} \ J of energy, out of which 1.126 \times 10^8 has been used to convert the water into steam.

Energy left = 10^{10} - 1.126 \times 10^8 = 9.88 \times 10^9 \ J

We use this energy to convert steam into vapors.

Q = \Delta E

Q = \Delta E = mc (T_{f} - T{i})\\T_{f} = \frac {\Delta E}{mc} + T_{i}\\ \\T_{f} = 100 + \frac{9.88 \times 10^{10}}{50 \times 1970}\\T_{f} = 100 + 10^8\\T_{f} = 10 ^{ \ 8} \ {^ \circ } C

With this temperature, we can easily interpret that the vapors will be dissociated in hydrogen and oxygen particles.

5 0
2 years ago
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