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2 years ago

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A car is travelling to the right with a speed of 42\,\dfrac{\text m}{\text s}42 s m 42, space, start fraction, m, divided by,

s, end fraction when the driver slams on the brakes. The car skids for 4.0\,\text s4.0s4, point, 0, space, s with constant acceleration before it comes to a stop. How many meters did the car skid before coming to a stop? Answer using a coordinate system where rightward is positive. Round the answer to two significant digits.

2 answers:

Effectus [21]2 years ago

7 0

Answer:

d = 84 m

Explanation:

As we know that when an object moves with uniform acceleration or deceleration then we can use equation of kinematics to find the distance moved by the object

here we know that

initial speed $v_i = 42 m/s$

final speed $v_f = 0$

time taken by the car to stop

$t = 4s$

now the distance moved by the car before it stop is given as

$d = \frac{v_f + v_i}{2} \times t$

now we have

$d = \frac{42 + 0}{2} \times 4$

$d = 84 m$

wolverine [178]2 years ago

5 0

Answer:

84

Explanation:

khan academy

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Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

MArishka [77]

Complete Question:

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Answer:

m = 0.001 M

For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/

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A nucleus whose mass is 3.499612×10^(−25) kg undergoes spontaneous alpha decay. The original nucleus disappears and there appear

Elanso [62]

Answer:

The sum of the kinetic energies of the alpha particle and the new nucleus = (6.5898 × 10⁻¹³) J

Explanation:

Old nucleus ---> New nucleus + alpha particle.

We will use the conservation of energy theorem for extremely small particles,

Total energy before split = total energy after split

That is,

Total energy of the original nucleus = (total energy of the new nucleus) + (total energy of the alpha particle)

Total energy of these subatomic particles is given as equal to (rest energy) + (kinetic energy)

Rest energy = mc² (Einstein)

Let Kinetic energy be k

Kinetic energy of original nucleus = k₀ = 0 J

Kinetic energy of new nucleus = kₙ

Kinetic energy of alpha particle = kₐ

Mass of original nucleus = m₀ = (3.499612 × 10⁻²⁵) kg

Mass of new nucleus = mₙ = (3.433132 × 10⁻²⁵) kg

Mass of alpha particle = mₐ = (6.640678 × 10⁻²⁷) kg

Speed of light = c = (3.0 × 10⁸) m/s

Total energy of the original nucleus = m₀c² (kinetic energy = 0, since it was originally at rest)

Total energy of new nucleus = (mₙc²) + kₙ

Total energy of the alpha particle = (mₐc²) + kₐ

(m₀c²) = (mₙc²) + kₙ + (mₐc²) + kₐ

kₙ + kₐ = (m₀c²) - [(mₙc²) + (mₐc²)

(kₙ + kₐ) = c² (m₀ - mₙ - mₐ)

(kₙ + kₐ) = (3.0 × 10⁸)² [(3.499612 × 10⁻²⁵) - (3.433132 × 10⁻²⁵) - (6.640678 × 10⁻²⁷)]

(kₙ + kₐ) = (9.0 × 10¹⁶)(0.00007322 × 10⁻²⁵) = (6.5898 × 10⁻¹³) J

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2 years ago

Hooke's law describes an ideal spring. Many real springs are better described by the restoring force (F Sp ) s =−kΔs−q(Δs) 3 (p)

Montano1993 [528]

Answer with Explanation:

We are given that

Restoring force, $(FS_p)s=-k\Delta s-q(\Delta s)^3$

$k=350N/m$

$q=750 N/m^3$

We have to find the work must you do to compress this spring 15 cm.

$\Delta s=15 cm=0.15 m$

Using 1 m=100 cm

Work done= $\int_{0}^{0.15}-Fd(\Delta s)$

W= $-\int_{0}^{0.15}(-k\Delta s-q(\Delta s)^3))d(\Delta s)$

$W=k[\frac{(\Delta s)^2}{2}]^{0.15}_{0}+q[\frac{(\Delta s)^4}{4}]^{0.15}_{0}$

$W=0.01125k+0.000127q=0.01125\times 350+0.000127\times 750$

$W=4.033 J$

Ideal spring work= $0.5k(\Delta s)^2=0.5\times 350\times (0.15)^2=3.938 J$

Percentage increase in work= $\frac{4.033-3.938}{3.928}\times 100=2.4$ %

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2 years ago

The table shows information about four students who are running around a track. Which statement is supported by the information

Vikentia [17]

Answer:

<em>Correct option: Mohammed has less kinetic energy than Autumn.</em>

Explanation:

<u>Kinetic Energy</u>

Is the energy an object has due to its motion. If the object has a mass m and travels at a speed v, then the kinetic energy K is:

$\displaystyle K=\frac{1}{2}mv^2$

The information about four students includes their mass and velocity as follows:

Autumn has a mass of m1=50 kg and a velocity (magnitude) of v1=4 m/s, thus their kinetic energy is:

$\displaystyle K_1=\frac{1}{2}50\cdot 4^2$

$K_1=400\ J$

Mohammed has a mass of m2=57 kg and a velocity (magnitude) of v2=3 m/s, thus their kinetic energy is:

$\displaystyle K_2=\frac{1}{2}57\cdot 3^2$

$K_2=256.5\ J$

Lexy has a mass of m3=53 kg and a velocity (magnitude) of v3=2 m/s, thus their kinetic energy is:

$\displaystyle K_3=\frac{1}{2}53\cdot 2^2$

$K_3=106\ J$

Chiang has a mass of m4=64 kg and a velocity (magnitude) of v4=5 m/s, thus their kinetic energy is:

$\displaystyle K_4=\frac{1}{2}64\cdot 5^2$

$K_4=800\ J$

Sorted from lower kinetic energy to higher:

Lexy, Mohammed, Autumn, Chiang. Thus:

Autumn has more kinetic energy than Chiang. False

Mohammed has less kinetic energy than Autumn. True

Lexy has more kinetic energy than Mohammed. False

Chiang has less kinetic energy than Lexy. False

Correct option: Mohammed has less kinetic energy than Autumn.

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2 years ago

Read 2 more answers

An all female guitar septet is getting ready to go on stage. The lead guitarist, Kira,who is always in tune, plucks her low E st

matrenka [14]

Answer:

Buffy > Aiko > Chandra > Freja > Evita

Explanation:

Beats occur when two waves of nearby frequencies overlap. The number of beats per second is equal to the difference in the frequency.

Let

$f=$ Frequency of Kira

$f_1=$ Frequency of Aiko

The beat frequency of Kira and Aiko is therefore,

$f_{beat}=|f_2-f_1|$

Substituting $f$ for $f_2$ and $3Hz$ for $f_{beat}$ , we get

$3Hz=|f-f_1|\\f_1=f+3Hz$

From the above equation, we see that beats increase

when

$f_2 =$ Frequency of Chandra

The beat frequency of Kira and Chandra is therefore,

$1Hz=|f-f_2|\\f_2=f-1Hz$

From the above equation, we see that beats decrease

When

$f_3=$ Frequency of Evita

The beat frequency of Kira and Evita is therefore,

$5Hz=|f-f_3|\\f_3=f-5Hz$

From the above equation, we see that beats decrease

When

$f_4=$ Frequency of Freja

The beat frequency of Kira and Freja is therefore,

$3Hz=|f-f_4|\\f_4=f-3Hz$

From the equation above, we see that beats decrease

When

$f_5=$ Frequency of Buffy

The beat frequency of Kira and Buffy becomes,

$4Hz=|f-f_5|\\f_5=f+4Hz$

From the equation above, we see that beats increase

Hence, the rank of members based on initial frequencies from largest to smallest is Buffy > Aiko > Chandra > Freja > Evita

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1 year ago