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photoshop1234 [79]
3 years ago
9

A force vector has a magnitude of 579 newtons and points at an angle 38.0° of below the positive x axis. (a) What is the the x

scalar component of the vector?
Fx = _____ N

(b) What is the y scalar component of the vector?
Fy = ______ N
Physics
1 answer:
m_a_m_a [10]3 years ago
8 0

Answer:

a) x- component of force = 432.62 N

b)y- component of force = 338 N

Explanation:

given,

magnitude of force = 549 N

angle between the force vector with positive horizontal x-axis(∅) = 38.0°

a) x-scalar component of the force vector

       x- component of force = F cos ∅

       x- component of force = 549 cos 38°

      x- component of force = 432.62 N

b) y-scalar component of the force vector

       y- component of force = F sin ∅

       y- component of force = 549 sin 38°

      y- component of force = 338 N

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A stone is thrown vertically into the air at an initial velocity of 96 ft/s. On Mars, the height s (in feet) of the stone above
vladimir1956 [14]

Answer:

240 ft

Explanation:

t = Time taken

u = Initial velocity = 96 ft/s

v = Final velocity

s = Displacement

a = Acceleration = 12 m/s² on Mars 32 ft/s² on Earth negative due to upward direction

Mars

s=ut+\frac{1}{2}at^2\\\Rightarrow s=96\times t+\frac{1}{2}\times -12\times t^2\\\Rightarrow s=96t-6t^2\ ft

Earth

s=ut+\frac{1}{2}at^2\\\Rightarrow s=96\times t+\frac{1}{2}\times -32\times t^2\\\Rightarrow s=96t-16t^2\ ft

Differentiating the first equation with respect to time we get

\frac{ds}{dt}=96-12t

Equating with zero

0=96-12t\\\Rightarrow t=\frac{96}{12}=8\ s

Differentiating the second equation with respect to time we get

\frac{ds}{dt}=96-32t

Equating with zero

0=96-32t\\\Rightarrow t=\frac{96}{32}=3\ s

Applying the time taken to the above equations, we get

s=96t-6t^2\ ft\\\Rightarrow s=96\times 8-6\times 8^2\\\Rightarrow s=384

s=96t-16t^2\\\Rightarrow s=96\times 3-16\times 3^2\\\Rightarrow s=144

Difference in height = 384-144 = 240 ft

The stone will travel 240 ft higher on Mars

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3 years ago
In 4 -5 sentences, summarize the science of the sport of curling and how it relates to friction
zalisa [80]
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3 years ago
A railroad freight car, mass 15,000 kg, is allowed to coast along a level track at a speed of 2.0 m/s. It collides and couples w
gayaneshka [121]

Answer:

The speed of the two cars after coupling is 0.46 m/s.

Explanation:

It is given that,

Mass of car 1, m₁ = 15,000 kg

Mass of car 2, m₂ = 50,000 kg

Speed of car 1, u₁ = 2 m/s

Initial speed of car 2, u₂ = 0

Let V is the speed of the two cars after coupling. It is the case of inelastic collision. Applying the conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V

V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}

V=\dfrac{15000\ kg\times 2\ m/s+0}{(15000\ kg+50000\ kg)}  

V = 0.46 m/s

So, the speed of the two cars after coupling is 0.46 m/s. Hence, this is the required solution.          

3 0
3 years ago
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Answer:

speed and velocity

Explanation:

8 0
2 years ago
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A cart, which has a mass of 2.30 kg is sitting at the top of an inclined plane, which is 4.50 meters long and meets the horizont
expeople1 [14]

Answer:

a) The gravitational potential energy before the cart rolls down the incline is 24.6 J.

b) The magnitude of the force that causes the cart to roll down is 5.47 N.

c) The acceleration of the cart is 2.38 m/s²

d) It takes the cart 1.94 s to reach the bottom of the incline.

e) The velocity of the cart at the bottom of the inclined plane is 4.62 m/s.

f) The kinetic energy of the cart as it reaches the bottom of the incline is 24.6 J.

g) The work done by the gravitational force is 24.6 J.

Explanation:

Hi there!

a) The gravitational potential energy is calculated using the following equation:

EP = m · g · h

Where:

EP = gravitational potential energy.

m = mass of the object.

g = acceleration due to gravity.

h = height at which the object is located.

The height of the inclined plane can be calculated using trigonomoetry:

sin 14.0° = height / lenght

sin 14.0° = height / 4.50 m

4.50 m · sin 14.0° = height

height = 1.09 m

Then, the gravitational potential energy will be:

EP = m · g · h

EP = 2.30 kg · 9.81 m/s² · 1.09 m = 24.6 J

The gravitational potential energy before the cart rolls down the incline is 24.6 J.

b) Please, see the attached figure for a graphical description of the problem and the forces acting on the cart. The force that causes the cart to accelerate down the incline is the horizontal component of the weight (Fwx in the figure). The magnitude of this force can be obtained using trigonometry:

sin 14° = Fwx / Fw

The weight of the cart (Fw) is calculated as follows:

Fw = m · g

Fw = 2.30 kg · 9.81 m/s²

Fw = 22.6 N

Then, the x-component of the weight will be:

FW · sin 14° = Fwx

22.6 N · sin 14° = Fwx

Fwx = 5.47 N

The magnitude of the force that causes the cart to roll down is 5.47 N.

c)Using the equation of Fwx we can calculate the acceleration of the cart:

Fwx = m · a

Where "m" is the mass of the cart and "a" is the acceleration.

Fwx / m = a

5.47 N / 2.30 kg = a

a = 2.38 m/s²

The acceleration of the cart is 2.38 m/s²

d) To calculate the time it takes the cart to reach the bottom of the incline, let´s use the equation of position of the cart:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the cart at time t.

x0 = initial position.

v0 = initial velocity.

a = acceleration.

t = time.

Considering the initial position as the point at which the cart starts rolling (x0 = 0) and knowing that the cart starts from rest (v0 = 0), let´s find the time it takes the cart to travel the 4.50 m of the inclined plane:

x = 1/2 · a · t²

4.50 m = 1/2 · 2.38 m/s² · t²

2 · 4.50 m / 2.38 m/s² = t²

t = 1.94 s

It takes the cart 1.94 s to reach the bottom of the incline.

e) The velocity of the cart at the bottom of the inclined plane can be obtained using the following equation:

v = v0 + a · t

v = 0 m/s + 2.38 m/s² · 1.94 s

v = 4.62 m/s

The velocity of the cart at the bottom of the inclined plane is 4.62 m/s.

f) The kinetic energy can be calculated using the following equation:

KE = 1/2 · m · v²

Where:

KE =  kinetic energy.

m = mass of the cart.

v = velocity of the cart.

KE = 1/2 · 2.30 kg · (4.62 m/s)²

KE = 24.6 J

The kinetic energy of the cart as it reaches the bottom of the incline is 24.6 J.

The gain of kinetic energy is equal to the loss of gravitational potential energy.

g) The work done by the gravitational force can be calculated using the work-energy theorem: the work done by the gravitational force is equal to the negative change in the gravitational potential energy:

W = -ΔPE

W = -(final potential energy - initial potential energy)

W = -(0 - 24.6 J)

W = 24.6 J

This can also be calculated using the definition of work:

W = Fw · d

Where "d" is the distance traveled in the direction of the force, that is the height of the inclined plane:

W = 22.6 N · 1.09 m = 24.6 J.

The work done by the gravitational force is 24.6 J.

4 0
4 years ago
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