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2 years ago

A 61,000-kg Rocky Mountain Double big rig is traveling at 0.5 m/s.

Physics

1 answer:

Firlakuza [10]2 years ago

5 0

Answer:

41.78 m/s

Explanation:

Momentum the defined as the product of mass and velocity of an object hence expressed as p=mv where p represent momentum, m is mass while v is velocity.

The law of conseevation of linear momentum states that the sum of the initial and final momentum should be equal. Therefore, in this case, MV=mu where M and m represent mass of 61, 000 kg and 730 respectively while v and u are their respective velocities.

Making u the subject then

$u=\frac {MV}{m}$

Substituting 0.5 m/s for V, 61000 kg for M and 730 kg for m then the velocity will be

$u=\frac {61000\times 0.5}{730}=41.780821917808 m/s\\u\approx 41.78 m/s$

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A professor's office door is 0.89 m wide, 2.0 m high, and 4.0 cm thick; has a mass of 25 kg ; and pivots on frictionless hinges.

taurus [48]

In order to answer this question ... strange as it may seem ...
we only need one of those measurements that you gave us
that describe the door.

The door is hanging on frictionless hinges, and there's a torque
being applied to it that's trying to close it. All we need to do is apply
an equal torque in the opposite direction, and the door doesn't move.

Obviously, in order for our force to have the most effect, we want
to hold the door at the outer edge, farthest from the hinges. That
distance from the hinges is the width of the door ... 0.89 m.

We need to come up with 4.9 N-m of torque,
applied against the mechanical door-closer.

Torque is (force) x (distance from the hinge).

4.9 N-m = (force) x (0.89 m)

Divide each side by 0.89m: Force = (4.9 N-m) / (0.89 m)

= 5.506 N .

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2 years ago

: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop

Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒ $Fa*sin(50)-Fb*sin(20)=0$

⇒ $Fa*sin(50)=Fb*sin(20)$

⇒ $Fa=2.24Fb$

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950

⇒ $Fa*cos(20)+Fb*cos(50)=950$

⇒ $2.24*Fb*cos(20)+Fb(50)=950$

⇒ $Fb*(2.24*cos(20)+cos(50))=950$

⇒ $Fb=\frac{950}{2.24*cos(20)+cos(50)}$

⇒ $Fb=\frac{950}{2.24*0.94+0.64}$

⇒ $Fb=\frac{950}{2.75}=345.5$

⇒ $Fa=2.24*Fb$

$=2.24*345.5$

$=773.93$

Therefore approximately, Fa=774 N and Fb=346 N

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2 years ago

A mercury thermometer has a glass bulb of interior volume 0.100 cm3 at 10°c. the glass capillary 10) tube above the bulb has an

Nadya [2.5K]

Initial volume of mercury is
V = 0.1 cm³

The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.

Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³

The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm

Answer: 4.5 cm

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2 years ago

Read 2 more answers

Imagine that the above hoop is a tire. the coefficient of static friction between rubber and concrete is typically at least 0.9.

Stels [109]

The hoop is attached.

Consider that the friction force is given by:
F = μ·N
 = μ·m·g·cosθ

We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·g·sinθ

Therefore:
-(1/2)·m·g·sinθ + m·g·sinθ = μ·m·g·cosθ
(1/2)·m·g·sinθ = μ·m·g·cosθ
μ = (1/2)·m·g·sinθ / m·g·cosθ
 = (1/2)·tanθ

Now, solve for θ:
θ = tan⁻¹(2·μ)
 = tan⁻¹(2·0.9)
 = 61°

Therefore, the maximum angle you could ride down without worrying about skidding is 61°.

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2 years ago

A car travels three-quarters of the way around a circle of radius 20.0 m in a time of 3.0 s at a constant speed. the initial vel

schepotkina [342]

20.3 divided by 3.0 will get u velocity and v times 3.0s

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2 years ago