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vladimir2022 [97]
3 years ago
15

While driving a 2150 kg car, carl steps on the gas pedal, accelerating at 4.0 m/s2. If the coefficient of friction between his c

ar and the road is
.25, how hard is the engine pulling forward?
a
13867.5 N
7417.5 N
9137.5 N
3332.5 N
Physics
1 answer:
Mrac [35]3 years ago
6 0

Answer:

d. 3332.5 [N]

Explanation:

To solve this problem we will use newton's second law, which tells us that the sum of forces is equal to the product of mass by acceleration.

Here we have two forces, the force that pushes the car to move forward and the friction force.

The friction force is equal to the product of the normal force by the coefficient of friction.

f = N * μ

f = (m*g) * μ

where:

N = weight of the car = 2150*9.81 = 21091.5 [N]

μ = 0.25

f = (21091.5) * 0.25

f = 5273 [N]

Now as the car is moving forward, the car wheels move clockwise. The friction force between the wheels of the car and the pavement must be counterclockwise, i.e. counterclockwise. Therefore the direction of this force is forward. This way we have:

F + f = m*a

F + 5273 = 2150*4

F = 8600 - 5273

F = 3327 [N]

Therefore the answer is d.

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Legacy issues $570,000 of 8.5%, four-year bonds dated January 1, 2019, that pay interest semiannually on June 30 and December 31
choli [55]

Answer:

1) Determine the total bond interest expense to be recognized.

Total bond interest expense over life of bonds:

Amount repaid:    

8 payments of $24,225:           $193,800    

Par value at maturity:                 $570,000    

Total repaid:                                   $763800 (193,800 + 570,000)  

Less amount borrowed:         $508050    

Total bond interest expense: $255750 (763800 - 508,050)

2)Prepare a straight-line amortization table for the bonds' first two years.

Semiannual Interest Period­ End; Unamortized Discount; Carrying Value

01/01/2019                                      61,950                           508,050  

06/30/2019                                      54,206                          515,794  

12/31/2019                                       46,462                         523,538  

06/30/2020                                       38,718                        531,282  

12/31/2020                                         30,974                          539,026

3) Record the interest payment and amortization on June 30:

June 30            Bond interest expense, dr                         31969  

                       Discount on bonds payable, Cr     (61950/8)  7743.75

                                        Cash, Cr                     ( 570000*8.5%/2)  24225  

4) Record the interest payment and amortization on December 31:

Dec 31                 Bond interest expense, Dr               31969  

                           Discount on bonds payable, Cr  7744  

                                    Cash, Cr                                24225

6 0
3 years ago
A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm
bogdanovich [222]

Answer:

A) 2.75 m/s  B) 0.1911 m    C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = \sqrt\frac{k}{M}

                                                       = \sqrt\frac{21}{0.1}

                                                       = 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic  Motion is given as:

                                x = A sin (\omega t + \phi)\\

Differentiating this to find speed of the block immediately before the collision:

                    v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\

As bullet strikes at equilibrium position so,

                                  φ = 0

                                   t= 2nπ

                             ⇒ cos (ω₀t + φ) = 1

                             ⇒ v= A\omega_{o}

                                       v=(.19)(14.49)\\v= 2.75 ms^{-1}

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

                              x= Bsin(\omega t + \phi)

To find B, consider law of conservation of energy

K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2}  \\P.E = \frac{1}{2} kB^{2}

\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec

5 0
3 years ago
The speed of sound in air is 10 times faster than the speed of a wave on a certain string. The density of the string is 0.002kg/
nata0808 [166]

Answer:

The tension on the string is 2.353 N.

Explanation:

Given;

the speed of sound in air, v₀ = 343 m/s

then, the speed of sound on the string, v = 343 / 10 = 34.3 m/s

mass per unit length, m/l = μ = 0.002 kg/m

The speed of sound on the string is given as;

v = \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu

where;

T is the tension on the string

T = (34.3)²(0.002)

T = 2.353 N

Therefore, the tension on the string is 2.353 N.

3 0
3 years ago
A child sits on the edge of a spinning merry-go-round that has a radius of 1.2 m. The child's speed is 5 m/s. What is the child'
Leya [2.2K]

Answer:

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Explanation:

The centripetal acceleration of the child, which is moving by uniform circular motion, is given by

a=\frac{v^2}{r}

where

v = 5 m/s is the tangential speed of the child

r = 1.2 m is the radius of the circular path

Substituting the numbers into the equation, we find

a=\frac{(5 m/s)^2}{1.2 m}=20.8 m/s^2

3 0
3 years ago
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melomori [17]
True, because being a competitive person means that you like challenging yourself to do more things, but if you’re really competitive you might challenge yourself too hard, and might hurt yourself, which will make you not want to really compete with anything anymore
4 0
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