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2 years ago

If a 2 cm vector represents speed, and 1 cm equals 5 m/s, how fast is the represented

Physics

1 answer:

Ne4ueva [31]2 years ago

6 0

Answer: 10m/s

Explanation: Since a single vector with length of 1cm expresses 5m/s, the vector which has a length doubled from the original vector should have the speed which is also doubled.

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Write the equation of a function h(t) that represents the amount of heat in joules required to heat the bar to a temperature of

bearhunter [10]

The initial temperature of the bar is 25. To get to the t temperature you need to add (t-25) degrees Celsius.

for 1 degree................... 7 Joules
y given degree........ p Joules

p=7y

In our case y=(t-25) .

h(t) = 7(t-25) which is the final answer.

8 0

2 years ago

What does Newton's 2nd Law explain?

kow [346]

Answer:

C. Why you must push harder to move a car farther.

Explanation:

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

$Acceleration = \frac {Force}{mass}$

Hence, Newton's 2nd Law explains why you must push harder to move a car farther because of its mass. Thus, it is important to increase the force that the engine provides and decrease the mass of the car.

6 0

2 years ago

A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp

Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

=> $F_x = Tcos \theta$

Substituting values

$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

4 0

3 years ago

Which statement best describes how the distance from a large river affects an area’s climate in the summer?

Montano1993 [528]

Answer:

Regions near rivers have water surfaces that rapidly change in temperature from cold to hot.

you have your own answer i only selected which is suitable for me . none of them is wrong they re excellent

3 0

2 years ago

Read 2 more answers

The rotating loop in an AC generator is a square 10.0 cm on each side. It is rotated at 60.0 Hz in a uniform field of 0.800 TO.

Genrish500 [490]

Answer:

Explanation:

Given a square side loop of length 10cm

L=10cm=0.1m

Then, Area=L²

Area=0.1²

Area=0.01m²

Given that, frequency=60Hz

And magnetic field B=0.8T

a. Flux Φ

Flux is given as

Φ=BA Sin(wt)

w=2πf

Φ=BA Sin(2πft)

Φ=0.8×0.01 Sin(2×π×60t)

Φ=0.008Sin(120πt) Weber

b. EMF in loop

Emf is given as

EMF= -N dΦ/dt

Where N is number of turns

Φ=0.008Sin(120πt)

dΦ/dt= 0.008×120Cos(120πt)

dΦ/dt= 0.96Cos(120πt)

Emf=-NdΦ/dt

Emf=-0.96NCos(120πt). Volts

c. Current induced for a resistance of 1ohms

From ohms law, V=iR

Therefore, Emf=iR

i=EMF/R

i=-0.96NCos(120πt) / 1

i=-0.96NCos(120πt) Ampere

d. Power delivered to the loop

Power is given as

P=IV

P=-0.96NCos(120πt)•-0.96NCos(120πt)

P=0.92N²Cos²(120πt) Watt

e. Torque

Torque is given as

τ=iL²B

τ=-0.96NCos(120πt)•0.1²×0.8

τ=-0.00768NCos(120πt) Nm

8 0

2 years ago