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3 years ago

PLEASE HELP!!!!

Physics

2 answers:

sergeinik [125]3 years ago

8 0

Answer:

Explanation:

ValentinkaMS [17]3 years ago

4 0

Answer: C

Explanation:

The answer is choice C.

The sum of the vectors is the resultant vector, which is where the net force is directed.

An example would be if you had a ball rolling on a table and you bumped the ball perpendicular to its initial velocity, then the ball would move at a diagonal angle rather than move straight in the direction where you bumped it.

Acceleration is the change in velocity over time, so the acceleration vector tells us how the velocity's direction is changing.

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Edit question A 37-cm-long wire of linear density 18 g/m vibrating at its second mode, excites the third vibrational mode of a t

Pachacha [2.7K]

Answer:

176.9N

Explanation:

The following data were given

wire length,L=37cm=0.37m

linear density=18g/m

tube length,=192cm=1.92m,

speed of sound,v=343m/s

Since it is an open-closed tube, the second harmonic frequency is expressed as

$f_{3}=3(\frac{v}{4l} )\\f_{3}=3(\frac{343}{4*1.92})\\f_{3}=133.98Hz$

The relationship between the tension, linear density and second harmonic frequency is expressed as

$f_{3}=\frac{1}{2l_{w}}\sqrt{\frac{T}{\alpha } } \\T=(f_{3}*2l_{w})^{2}\alpha \\T=(133.984*2*0.37)^{2}*18*10^{-3}\\T=176.9N$

7 0

3 years ago

To answer this question, suppose that each vehicle is moving at 7.69 m/s and that they undergo a perfectly inelastic head-on col

Maurinko [17]

Answer:

Force(F) = -80,955.01 N

Explanation:

We need to first determine the impulse that the truck driver received from the car during the collision

So; m₁v₁ - m₂v₂ = (m₁m₂)v₀

where;

m₁ = mass of the truck = 4280 kg

v₁ = v₂ = speed of the each vehicle = 7.69 m/s

m₂ = mass of the car = 810 kg

Substituting our data; we have:

(4280×7.69) - (810×7.69) = (4280+810)v₀

32913.2 - 6228.9 = (5090)v₀

26684.1 = (5090)v₀

v₀ = $\frac{26684.1}{5090}$

v₀ = 5.25 m/s

NOW, Impulse on the truck = m (v₀ - v)

= 4280 × (5.25 - 7.69)

= 4280 × (-2.44)

= -10,443.2 kg. m/s

Force that the seat belt exert on the truck driver can be calculated as:

Impulse = Force × Time

-10,443.2 kg. m/s = F (0.129)

F = $\frac{-10,443.2}{0.129}$

Force(F) = -80,955.01 N

Thus, the Force that the seat belt exert on the truck driver = -80,955.01 N

4 0

4 years ago

On a hot day, the deck of a small ship reaches a temperature of 48

AlekseyPX

The final temperature of the seawater-deck system is 990°C.

The increment in temperature adds up the thermal energy into the object. This energy is Heat energy.

The deck of a small ship reaches a temperature Ti= 48.17°C seawater on the deck to cool it down. During the cooling, heat Q =3,710,000 J are transferred to the seawater from the deck. Specific heat of seawater= 3,930 J/kg°C.

Suppose for 1 kg of sea water, the heat transferred from the system is given by

3,710,000 = 1 x 3,930 x (T - 48.17)

T = 990°C to the nearest tenth.

The final temperature of the seawater-deck system is 990°C.

Learn more about heat.

brainly.com/question/13860901

#SPJ1

6 0

2 years ago

Can someone help me with 1-2

ira [324]

Can't really plot a graph here for question 1.

2a) The car speeds up from A to B. The car travels at a constant speed from B to C. The car slows down to a stop from C to D.

b) From the graph, at 10 seconds, the car is moving at 20 m/s.

4 0

3 years ago

Which of the following represents a possible magnitude for the force of static friction when Xavier applied 72.1 Newtons of forc

lana66690 [7]

The possible magnitude for the force of static friction on the stationary cart is 72.1 N.

The given parameters:

<em>Applied force on the cart, F = 72.1 N</em>

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Based on Newton's second law of motion, the force applied to object is directly proportional to the product of mass and acceleration of the object.

F = ma

Static frictional force is the force resisting the motion of an object at rest.

$\Sigma F = 0\\\\F -F_f = 0$

where;

$F_f$ is the frictional force

$F= F_f \\\\72.1 = F_f\\\\F_f = 72.1\ N$

Thus, the possible magnitude for the force of static friction on the stationary cart is 72.1 N.

Learn more about Newton's second law of motion: brainly.com/question/25307325

8 0

2 years ago