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mafiozo [28]
2 years ago
7

A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, take

n as the xy plane. The 4.20-kg puck has a velocity of 1.60î m/s at one instant. Eight seconds later, its velocity is (6.00î + 4.0ĵ) m/s. (a) Assuming the rocket engine exerts a constant horizontal force, find the components of the force.

Physics
1 answer:
adell [148]2 years ago
6 0

Answer:

F_{x}=2.31N to the right.

F_{y}=2.1N to in the upwards direction.

Explanation:

In order to solve this problem, we must first start by drawing a diagram of the situation. (See attached diagram).

So, remember that a force is determined by multiplying the mass of the parcticle by its acceleration:

F=ma

so in order to find the components of the force, we need to start by finding its acceleration.

Acceleration is found by using the following formula:

a=\frac{V_{f}-V{0}}{t}

so we can subtract the two vectors, like this:

a=\frac{(6.00i+4.0j)m/s-1.60i m/s}{8s}

which yields:

a=\frac{(4.4i+4.0j)m/s}{8s}

or:

a=(0.55i + 0.5j) m/s^{2}

so now I can find the components of the force:

F=(4.2kg)(0.55i + 0.5j) m/s^{2}

which yields:

F=(2.31i+2.1j)N

so the components of the force are:

F_{x}=2.31N to the right.

F_{y}=2.1N to in the upwards direction.

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A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between
Oduvanchick [21]

Answer:

Net electric field, E_{net}=91406.24\ N/C

Explanation:

Given that,

Charge 1, q_1=7.5\ nC=7.5\times 10^{-9}\ C

Charge 2, q_2=-2.9\ nC=-2.9\times 10^{-9}\ C

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

E_1=\dfrac{kq_1}{r^2}

E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}

E_1=65917.96\ N/C

Electric field due to charge 2 is given by :

E_2=\dfrac{kq_2}{r^2}

E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}

E_2=25488.28\ N/C

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

E_{net}=E_1+E_2

E_{net}=65917.96+25488.28

E_{net}=91406.24\ N/C

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

3 0
2 years ago
an elastic cord 61 cm long when a weight of 75N hangs from it, but 85cm when a weight of 210N hangs from it. what is the spring
pishuonlain [190]

Answer:

560 N/m

Explanation:

F = kx

75 N = k (0.61 m − L)

210 N = k (0.85 m − L)

Divide the equations:

2.8 = (0.85 − L) / (0.61 − L)

2.8 (0.61 − L) = 0.85 − L

1.708 − 2.8L = 0.85 − L

0.858 = 1.8L

L = 0.477

Plug into either equation and find k.

75 = k (0.61 − 0.477)

k = 562.5

Rounded to two significant figures, k = 560 N/m.

3 0
2 years ago
Merry-go-rounds are a common ride in park playgrounds. The ride is a horizontal disk that rotates about a vertical axis at their
Vera_Pavlovna [14]

Answer:

A = 2.36m/s

B = 3.71m/s²

C = 29.61m/s2

Explanation:

First, we convert the diameter of the ride from ft to m

10ft = 3m

Speed of the rider is the

v = circumference of the circle divided by time of rotation

v = [2π(D/2)]/T

v = [2π(3/2)]/4

v = 3π/4

v = 2.36m/s

Radial acceleration can also be found as a = v²/r

Where v = speed of the rider

r = radius of the ride

a = 2.36²/1.5

a = 3.71m/s²

If the time of revolution is halved, then radial acceleration is

A = 4π²R/T²

A = (4 * π² * 3)/2²

A = 118.44/4

A = 29.61m/s²

7 0
2 years ago
3. You are walking in Paris alongside the Eiffel Tower and suddenly a croissant smacks you on the head and knocks you to the gro
Elza [17]

As absurd as the concept is, we must assume that a croissant
can fall 300.5 meters through the moisture-laden, perfumed and
polluted Parisian air with no air resistance whatsoever.

Acceleration due to gravity on Earth:  9.8 m/s²

Distance in clean,
unimpeded free-fall         = (1/2) (acceleration) x (time²) 

                                            300.5 m  =  (1/2)  (9.8 m/s²) (T²)
Divide each side
by (4.9 m/s²):                  (300.5 m) / (4.9 m/s²)  =  T²

Take the square root
of each side:                     T = √(300.5/4.9) (s²)

                                                               =  7.831 seconds .

5 0
2 years ago
|| Climbing ropes stretch when they catch a falling climber, thus increasing the time it takes the climber to come to rest and r
Otrada [13]

To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.

Newton's second law is defined as

F = ma

Where,

m = mass

a = acceleration

From this equation we can figure the acceleration out, then

a = \frac{F}{m}

a = \frac{11*10^3}{80}

a = 137.5m/s

From the cinematic equations of motion we know that

v_f^2-v_i^2 = 2ax

Where,

v_f =Final velocity

v_i =Initial velocity

a = acceleration

x = displacement

There is not Final velocity and the acceleration is equal to the gravity, then

v_f^2-v_i^2 = 2ax

0-v_i^2 = 2(-g)x

v_i =\sqrt{2gx}

v_i = \sqrt{2*9.8*4.8}

v_i = 9.69m/s

From the equation of motion where acceleration is equal to the velocity in function of time we have

a = \frac{v_i}{t}

t = \frac{v_i}{a}

t =\frac{9.69}{137.5}

t = 0.0705s

Therefore the time required is 0.0705s

4 0
2 years ago
Read 2 more answers
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