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2 years ago

Micah knows that a car had a change in velocity of 15 m/s. What does micah need to determine acceleration.

2 answers:

katrin2010 [14]2 years ago

4 0

The time component is needed. The acceleration is the change of velocity divided by the time in when this change of velocity happens.

Nadusha1986 [10]2 years ago

3 0

Answer:

He needs time which is B.

Explanation:

I took the test

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Your neighbor Paul has rented a truck with a loading ramp. The ramp is tilted upward at 25°, and Paul is pulling a large crate u

Alisiya [41]

Explanation:

It is given that,

The ramp is tilted upwards at 25 degrees and Paul is pulling a large crate up the ramp with a rope that angles 10° above the ramp.

Total angle with respect to ramp is 35 degrees.

If Paul pulls with a force of 550 N.

The horizontal component of the force is given by :

$F_x=F\ cos\theta$

$F_x=550\ cos(35)$

$F_x=450.53\ N$

The vertical component of the force is given by :

$F_y=F\ sin\theta$

$F_y=550\ sin(35)$

$F_y=315.46\ N$

Hence, this is the required solution.

4 0

3 years ago

An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.275 T. If the kinetic energy of the electr

xxMikexx [17]

Answer:

Radius, $r=2.14\times 10^{-5}\ m$

Explanation:

It is given that,

Magnetic field, B = 0.275 T

Kinetic energy of the electron, $E=4.9\times 10^{-19}\ J$

Kinetic energy is given by :

$E=\dfrac{1}{2}mv^2$

$v=\sqrt{\dfrac{2E}{m}}$

$v=\sqrt{\dfrac{2\times 4.9\times 10^{-19}}{9.1\times 10^{-31}}}$

v = 1037749.04 m/s

The centripetal force is balanced by the magnetic force as :

$qvB\ sin90=\dfrac{mv^2}{r}$

$r=\dfrac{mv}{qB}$

$r=\dfrac{9.1\times 10^{-31}\times 1037749.04}{1.6\times 10^{-19}\times 0.275 }$

$r=2.14\times 10^{-5}\ m$

So, the radius of the circular path is $2.14\times 10^{-5}\ m$ . Hence, this is the required solution.

3 0

3 years ago

Read 2 more answers

An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounce

Ierofanga [76]

Answer:

2.67 m

Explanation:

k = Spring constant = 1.5 N/m

g = Acceleration due to gravity = 9.81 m/s²

l = Unstretched length

Frequency of SHM motion is given by

$f_s=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

Frequency of pendulum is given by

$f_p=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$

Given in the question

$f_p=\dfrac{1}{2}f_s$

$\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow \sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{g}{l}=\dfrac{1}{4}\dfrac{k}{m}\\\Rightarrow l=\dfrac{4gm}{k}\\\Rightarrow l=\dfrac{4\times 9.81\times \dfrac{1}{9.81}}{1.5}\\\Rightarrow l=2.67\ m$