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nikitadnepr [17]
3 years ago
6

What is the equivalent of 0° C in Kelvin? Help will give brainiest

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

Answer:

273.15 dat is the answer im pretty sure

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Two students, standing on skateboards, are initially at rest, when they give each other a shove! After the shove, one student (7
Elden [556K]

Answer:

The other student (59kg) moves right at 7.44 m/s

Explanation:

Given;

mass of the first student, m₁ = 77kg

mass of the second student, m₂ = 59kg

initial velocity of the first student, u₁ = 0

initial velocity of the second student, u₂ = 0

final velocity of the first student, v₁ = 5.7 m/s left

final velocity of the second student, v₂ = ? right

Apply the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(77 x 0) + (59 x 0) = (-77 x 5.7) + (59 x  v₂)

0 = - 438.9 + 59v₂

59v₂ = 438.9

v₂ = 438.9 / 59

v₂ = 7.44 m/s to the right

Therefore, the other student (59kg) moves right at 7.44 m/s

6 0
3 years ago
56. What is the range of magnitudes that can be attained from the
Svetach [21]

Answer:

Explanation:

Given the magnitude of the forces, 7N and 2N, the minimum combining force acting on the forces are when the force's is acting in opposite direction.

Magnitude of the force in opposite direction is 7N - 5N = 2N

The maximum combining force occurs when they act in the same direction. Magnitude of the force in the same direction is 5N+7N = 12N

Hence the range of magnitude requires is 2N≤F≤12N

8 0
3 years ago
Suppose that a particular artillery piece has a range R = 9880 yards . Find its range in miles. Use the facts that 1mile=5280ft
Solnce55 [7]

Answer:

R = 9880 yd * 3 ft/yd / 5280 ft/mi = 5.61 mi

If you do it in steps

R = 9880 yd * 3 ft/yd = 29640 ft

R = 29640 ft / 5280 ft/mi = 5.61 mi

6 0
3 years ago
The correct order of the planets is:
Nastasia [14]
False its Jupiter than Saturn

6 0
3 years ago
A stone is dropped from the upper observation deck of a tower, 750 m above the ground. (assume g = 9.8 m/s2.) (a) find the dista
Gekata [30.6K]
(a) The stone moves by uniform accelerated motion, with constant acceleration g=9.81 m/s^2 directed downwards, and its initial vertical position at time t=0 is 750 m. So, the vertical position (in meters) at any time t can be written as
y(t)= y_0 -  \frac{1}{2}gt^2= 750 - 4.9 t^2

(b) The time the stone takes to reach the ground is the time at which the vertical position of the stone becomes zero: y(t)=0. So, we can write
750-4.9 t^2 = 0
from which we find the time t after which the stone reaches the ground:
t= \sqrt{\frac{750 m}{4.9 m/s^2 }}= 12.37 s

(c) The velocity of the stone at time t can be written as
v(t) = -gt
because it is an accelerated motion with initial speed zero. Substituting t=12.37 s, we find the final velocity of the stone:
v(12.37 s)=-(9.81 m/s^2)(12.37 s)=-121.3 m/s

(d) if the stone has an initial velocity of v_0 = 6 m/s, then its law of motion would be
y(t)=y_0 - v_0t -  \frac{1}{2}gt^2
and we can find the time it needs to reach the ground by requiring again y(t)=0:
0=750 - 6t - 4.9 t^2
which has two solutions: one is negative so we neglect it, while the second one is t=11.78 s, so this is the time after which the stone reaches the ground.

5 0
3 years ago
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