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ladessa [460]
3 years ago
5

A hammer slides down a roof that makes a 30.0°angle with the horizontal. What are the magnitudes of the components of the hammer

's velocity at the edge of the roof if it is moving at a speed of 8.25 m/s
Physics
1 answer:
dlinn [17]3 years ago
7 0
<h2>Answer:7.14ms^{-1},4.125ms^{-1}</h2>

Explanation:

Whenever an object is moving in a 2D frame,its motion can be analysed as if it is travelling in two independent 1D frames.

One of such independent 1D frames are along horizontal and another along vertical.

Let v be the total velocity.

Given that,v=8.25ms^{-1}

We call the horizontal velocity as v_{h} and the vertical velocity as v_{v}.

v_{h}=vCos\alpha

v_{v}=vSin\alpha

where \alpha is the angle between the object and horizontal.

It is given that \alpha =30^{0}

v_{h}=8.25\times Cos(30^{0})=7.14ms^{-1}

v_{v}=8.25\times Sin(30^{0})=4.125ms^{-1}

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A 0.405 kg mass is attached to a spring with a force constant of 26.3 N/m and released from rest a distance of 3.31 cm from the
miv72 [106K]

Answer:

0.231 m/s

Explanation:

m = mass attached to the spring = 0.405 kg

k = spring constant of spring = 26.3 N/m

x₀ = initial position = 3.31 cm = 0.0331 m

x = final position = (0.5) x₀ = (0.5) (0.0331) = 0.01655 m

v₀ = initial speed = 0 m/s

v = final speed = ?

Using conservation of energy

Initial kinetic energy + initial spring energy = Final kinetic energy + final spring energy

(0.5) m v₀² + (0.5) k x₀² = (0.5) m v² + (0.5) k x²

m v₀² + k x₀² = m v² + k x²

(0.405) (0)² + (26.3) (0.0331)² = (0.405) v² + (26.3) (0.01655)²

v = 0.231 m/s

8 0
3 years ago
State examples of a transverse wave. ​
laiz [17]

ripples on the surface of water.

vibrations in a guitar string.

a Mexican wave in a sports stadium.

electromagnetic waves – eg light waves, microwaves, radio waves.

seismic S-waves.

6 0
2 years ago
Read 2 more answers
Two 4.0 kg masses are 1.0 m apart on a frictionless table. Each has 1.0 μC of charge.What is the magnitude of the electric force
xeze [42]

Coulomb's law:

  Force = (<span>8.99×10⁹ N m² / C²<span>) · (charge₁) · (charge₂) / distance²

            = (</span></span><span>8.99×10⁹ N m² / C²<span>) (1 x 10⁻⁶ C) (1 x 10⁻⁶ C)  / (1.0 m)²

            = (8.99×10⁹ x 1×10⁻¹² / 1.0)  N

            =      8.99×10⁻³  N

            =        0.00899 N repelling.

Notice that there's a lot of information in the question that you don't need.
It's only there to distract you, confuse you, and see whether you know
what to ignore.

-- '4.0 kg masses';  don't need it. 
   Mass has no effect on the electric force between them.

-- 'frictionless table';  don't need it.
   Friction has no effect on the force between them,
only on how they move in response to the force.
</span></span>
7 0
3 years ago
The parallel plates in a capacitor, with a plate area of 8.00 cm2 and an air-filled separation of 2.70 mm, are charged by a 8.70
MrRa [10]

Answer:

a)  ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J,  d)  W = 150 10⁻¹² J

Explanation:

Let's find the capacitance of the capacitor

         C = \epsilon_o \frac{A}{d}

         C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³

         C = 2.62 10⁻¹² F

for the initial data let's look for the accumulated charge on the plates

          C = \frac{Q}{\Delta V}

          Q₀ = C ΔV

           Q₀ = 2.62 10⁻¹² 8.70

           Q₀ = 22.8 10⁻¹² C

a) we look for the capacity for the new distance

          C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³

          C₁ = 1.04 10⁻¹² F

       

          C₁ = Q₀ / ΔV₁

          ΔV₁ = Q₀ / C₁

          ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²

          ΔV₁ = 21.9 V

b) initial stored energy

          U₀ = \frac{Q_o}{ 2C}

          U₀ = (22.8 10⁻¹²)²/(2  2.62 10⁻¹²)

          U₀ = 99.2 10⁻¹² J

c) final stored energy

          U_f = (22.8 10⁻¹²) ² /(2  1.04 10⁻⁻¹²)

          U_f = 249.9 10⁻¹² J

d) the work of separating the plates

as energy is conserved work must be equal to energy change

          W = U_f - U₀

          W = (249.2 - 99.2) 10⁻¹²

          W = 150 10⁻¹² J

note that as the energy increases the work must be supplied to the system

6 0
2 years ago
Projectile motion 1 and 2​
shtirl [24]

1. Vf =62.57778m/s

h=62.5778

theta final=45°????

8 0
3 years ago
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