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JulsSmile [24]
3 years ago
11

A large chunk of ice breaks off an Alaskan glacier and lands in the ocean.

Physics
1 answer:
siniylev [52]3 years ago
8 0

Answer:

b hope this helps

Explanation:

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hockey puck slides across the ice with an initial velocity of 7.2 m/s. It has a deceleration of 1.1 m/s2 and is traveling toward
dimulka [17.4K]

For this use the formula:

d = Vo * t - (at^2) / 2

Clearing t:

t = d/(v + 0.5*a)

Replacing:

t = 5 m / (7.2 m/s + 0.5 * (-1.1 m/s²)

Resolving:

t = 5 m / (7.2 m/s + (-0.55 m/s²)

t = 5 m / 6.65 m/s

t = 0.75 s

Result:

The time will be <u>0.75 seconds.</u>

7 0
3 years ago
Read 2 more answers
Where do estuaries form?
vekshin1

Answer:

An estuary is the area where a river meets the sea or ocean, where fresh water from the river meets salt water from the sea.

7 0
3 years ago
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n alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.94 cm in a uniform magnetic field with B = 1.10 T.
9966 [12]

Answer:

a). V = 3.13*10⁶ m/s

b). T = 1.19*10^-7s

c). K.E = 2.04*10⁵

d). V = 1.02*10⁵V

Explanation:

q = +2e

M = 4.0u

r = 5.94cm = 0.0594m

B = 1.10T

1u = 1.67 * 10^-27kg

M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg

a). Centripetal force = magnetic force

Mv / r = qB

V = qBr / m

V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27

V = 2.09088 * 10^-20 / 6.68 * 10^-27

V = 3.13*10⁶ m/s

b). Period of revolution.

T = 2Πr / v

T = (2*π*0.0594) / 3.13*10⁶

T = 1.19*10⁻⁷s

c). kinetic energy = ½mv²

K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²

K.E = 3.27*10^-14J

1ev = 1.60*10^-19J

xeV = 3.27*10^-14J

X = 2.04*10⁵eV

K.E = 2.04*10⁵eV

d). K.E = qV

V = K / q

V = 2.04*10⁵ / (2eV).....2e-

V = 1.02*10⁵V

7 0
3 years ago
What type of magnetism is present in a magnet that you use to place a photograph on the refrigerator?
Vika [28.1K]

Answer:

C) Contact

Explanation:

The magnet requires almost direct <u>contact</u> with the fridge to start its magnetic properties.

7 0
4 years ago
Read 2 more answers
A 1.2kg stone is tied to a string and swung in a vertical circle with a radius of 0.75m. The string can withstand a tension of 4
san4es73 [151]

Hello!

We can begin by summing the forces acting on the stone when it is at the bottom of its trajectory.

Refer to the free-body diagram in the image below for clarification.

We have the force of tension (produced by the string) and the force of gravity acting in opposite directions, so:
\Sigma F = T - F_g

The net force is equivalent to the centripetal force experienced by the stone. Recall the equation for centripetal force for uniform circular motion:
F_c = \frac{mv^2}{r}

m = mass of object (1.2 kg)
v = velocity of object (? m/s)
r = radius of circle (0.75 m)

The centripetal force is the resultant of the forces of tension and gravity, and points upward (same direction as the tension force) since the tension force is greater.

Therefore:
\frac{mv^2}{r} = T - Mg

We can solve the equation for 'v':

mv^2 = r(T - Mg) \\\\v^2 = \frac{r(T - Mg)}{m}\\\\v = \sqrt{\frac{r(T - Mg)}{m}}

Plug in values and solve.

v = \sqrt{\frac{(0.75)(40 - 1.2(9.8))}{1.2}} = \boxed{4.201 \frac{m}{s}}

3 0
2 years ago
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