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3 years ago

8

A bike, a truck, and a train—all without passengers, motors, or engines—roll down the same hill. Put the vehicles in order from

the least amount of motion energy to the most.

1 answer:

Bezzdna [24]3 years ago

3 0

Answer:

Train Bike Truck

Explanation:

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A 200 g block is pressed against a spring of force constant 1.40 kN/m until the block compresses the spring 10.0 cm. The spring

tino4ka555 [31]

Answer:

Explanation:

This problem bothers on the energy stored in a spring in relation to conservation of energy

Given data

Mass of block m =200g

To kg= 200/1000= 0.2kg

Spring constant k = 1.4kN/m

=1400N/m

Compression x= 10cm

In meter x=10/100 = 0.1m

Using energy considerations or energy conservation principles

The potential energy stored in the spring equals the kinetic energy with which the block move away from the spring

Potential Energy stored in spring

P.E=1/2kx^2

Kinetic energy of the block

K.E =1/mv^2

Where v = velocity of the block

K.E=P.E (energy consideration)

1/2kx^2=1/mv^2

Kx^2= mv^2

Solving for v we have

v^2= (kx^2)/m

v^2= (1400*0.1^2)/0.2

v^2= (14)/0.2

v^2= 70

v= √70

v= 8.36m/s

a. Distance moved if the ramp exerts no force on the block

Is

S= v^2/2gsinθ

Assuming g= 9. 81m/s^2

S= (8.36)^2/2*9.81*sin60

S= 69.88/19.62*0.866

S= 69.88/16.99

S= 4.11m

6 0

3 years ago

How does Newton's second law apply to the egg drop?

avanturin [10]

<span> Newtons First Law is applied on my egg experiment because it will not move or change it's acceleration until a force acts upon it. In this case, one example of those forces would be Mr. Baker picking up the egg project. Newton's Second Law is applied because of the acceleration caused by natural forces as the egg is plummeting to the earth.</span>

4 0

3 years ago

At time t=0 a positively charged particle of mass m=3.57 g and charge q=9.12 µC is injected into the region of the uniform magne

larisa [96]

Answer:

10.78 s

Explanation:

The force on the charge is computed by using the equation:

$F^{\to}= qE^{\to} +q (v^{\to} + B^{\to}) \\ \\ F^{\to} = (9.12 \times 10^{-6}) *278 (-\hat k) +9.12 *10^{-6} *2.1 *0.18 (\hat i * \hat k) \\ \\ F^{\to} = -2.535 *10^{-3} \hat k -3.447*10^{-6} \hat j$

F = ma

∴

$a ^{\to}= \dfrac{F^{\to}}{m}$

$a ^{\to}= \dfrac{-1}{3.57\times 10^{-3}}(2.535*10^{-3}\hat k + 3.447*10^{-6} \hat j)$

$a ^{\to}=-0.710 \hat k -9.656*10^{-4} \hat j$

At time t(sec; the partiCle velocity becomes $v(t) = 3.78 v_o$

The velocity of the charge after the time t(sec) is expressed by using the formula:

$v^{\to}= v_{o \ \hat i} + a^{\to }t \\ \\ \implies (2.1)\hat i -0.710 t \hat k -9.656 \times 10^{-4} t \hat j = 3.78 v_o \\ \\ \implies (2.1)^2 +(0.710\ t)^2+ (9.656 *10^{-4}t )^2 = (3.78 *2.1^2 \\ \\ \implies 4.41 +0.5041 t^2 +9.324*10^{-7} t^2 = 63.012 \\ \\ \implies 4.41 +0.5041 t^2 = 63.012\\ \\ 0.5041t^2 = 63.012-4.41 \\ \\ t^2 = \dfrac{58.602}{0.5041} \\ \\ t^2 = 116.25 \\ \\ t = \sqrt{116.25} \\ \\ \mathbf{t = 10.78 \ s}$

8 0

3 years ago

The __________ lobe contains the area of the cortex involved in auditory processing called the primary __________ cortex. A. par

Vlada [557]

<span>The temporal lobe contains the area of the cortex involved in auditory processing called the primary auditory cortex.</span><span>

The temporal </span>lobe<span> is associated with </span>auditory processing<span> and olfaction.
</span>The primary auditory cortex<span> is the </span>part<span> of the temporal </span>lobe which<span> processes </span>auditory <span>information.</span>

8 0

3 years ago

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What is the net force on the loop from the wire if the loop carries a clockwise current i2 = 50 a?

natima [27]

WE are given
I2 = 50 A
The formula for the net force is
F = I2 L B
where I2 is the current
L is the length
and
B is the magnetic field
If we know L and B, we can substitute the values into the formula and solve fore the force

5 0

4 years ago