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egoroff_w [7]
3 years ago
11

Two masses (5.3kg and 7.5kg) are fastened together with a small amount of explosive. They are loaded into a spring gun that is t

ilted 60.0 ◦ above the horizontal and launched at a speed of 25m/s. At the top of the trajectory, the explosive detonates (assume the mass of explosive is negligble). After it detonates, the 5.3kg mass isn’t moving at all. What is the velocity of the second mass just after the explosion?
Physics
1 answer:
Vsevolod [243]3 years ago
8 0

Answer:

8.2

Explanation:

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8. When a 100 N bag of nails hangs motionless from a single vertical strand of rope, how many newtons of tension are exerted in
Svetllana [295]

If the bag is motionless, then it's not accelerating up or down.
That fact right there tells you that the net vertical force on it
is zero.  So the sum of any upward forces on it is exactly equal
to the downward gravitational force ... the bag's "weight".

If the bag is suspended from a single rope, then the tension
in the rope must be equal to the 100-N weight of the bag.

And if there are four ropes holding it up, then the sum of
the four tensions is 100N.  If the ropes have been carefully
adjusted to share the load equally, then the tension is 25N
in each rope.

8 0
3 years ago
Write the equation of a function h(t) that represents the amount of heat in joules required to heat the bar to a temperature of
bearhunter [10]
The initial temperature of the bar is 25. To get to the t temperature you need to add (t-25) degrees Celsius.

for 1 degree................... 7 Joules
      y given degree........  p Joules

p=7y

In our case y=(t-25) .

h(t) = 7(t-25) which is the final answer.

8 0
3 years ago
As a new electrical technician, you are designing a large solenoid to produce a uniform 0.170 T magnetic field near the center o
MrRa [10]

Answer:

18.6012339739 A

Explanation:

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

L = Length of wire = 55 cm

N = Number of turns = 4000

I = Current

Magnetic field is given by

B=\dfrac{\mu_0NI}{L}\\\Rightarrow I=\dfrac{BL}{\mu_0N}\\\Rightarrow I=\dfrac{0.17\times 0.55}{4\pi \times 10^{-7}\times 4000}\\\Rightarrow I=18.6012339739\ A

The current necessary to produce this field is 18.6012339739 A

7 0
3 years ago
Carlos gives a grocery cart a 60-N push. The cart has a mass of 40 kg. What is the cart's acceleration?
andre [41]
<h3>Answer:</h3>

1.5 m/s²

<h3>Explanation:</h3>

We are given;

Force as 60 N

Mass of the Cart as 40 kg

We are required to calculate the acceleration of the cart.

  • From the newton's second law of motion, the rate of change in momentum is directly proportional to the resultant force.
  • That is, F = ma , where m is the mass and a is the acceleration

Rearranging the formula we can calculate acceleration, a

a = F ÷ m

  = 60 N ÷ 40 kg

  = 1.5 m/s²

Therefore, the acceleration of the cart is 1.5 m/s²

3 0
3 years ago
A force of 5N and a force of 8N act to the same point and are inclined at 45degree to each other. Find the magnitude and directi
Alex_Xolod [135]
  • Magnitude: 12.1 N.
  • Direction: 17.0° to the 8 N force.
<h3>Explanation</h3>

Refer to the diagram attached (created with GeoGebra). Consider the 5 N force in two directions: parallel to the 8 N force and normal to the 8 N force.

  • \displaystyle F_{\text{1, Parallel}} = F_1 \cdot \cos{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}.
  • \displaystyle F_{\text{1, Normal}} = F_1 \cdot \sin{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}.

The sum of forces on each direction will be the resultant force on that direction:

  • Resultant force parallel to the 8 N force: (8 + \dfrac{5\sqrt{2}}{2})\;\text{N}.
  • Resultant force normal to the 8 N force: \dfrac{5\sqrt{2}}{2}\;\text{N}.

Apply the Pythagorean Theorem to find the magnitude of the resultant force.

\displaystyle \Sigma F = \sqrt{{(8 + \dfrac{5\sqrt{2}}{2})}^2 + {(\dfrac{5\sqrt{2}}{2})}^2} = 12.1\;\text{N} (3 sig. fig.).

The size of the angle between the resultant force and the 8 N force can be found from the tangent value of the angle. Tangent of the angle:

\displaystyle \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}} = \dfrac{8 + \dfrac{5\sqrt{2}}{2}}{\dfrac{5\sqrt{2}}{2}} \approx 0.306491.

Find the size of the angle using inverse tangent:

\displaystyle \arctan{ \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}}} = \arctan{0.306491} = 17.0\textdegree.

In other words, the resultant force is 17.0° relative to the 8 N force.

4 0
3 years ago
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