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3 years ago

A car traveling at 37m/s starts to decelerate steadily. It comes to a complete stop in 15 seconds. What is it’s acceleration

Physics

1 answer:

LuckyWell [14K]3 years ago

6 0

<u>Answer</u>

The acceleration is

$a=-2.5ms^{-2}$ to the nearest tenth

<u>Explanation</u>

Since the car was travelling at $37ms^{-1}$ before it starts to decelerate, the initial velocity is

$u=37ms^{-1}$ .

The final velocity is $v=0ms^{-1}$ , because the car came to a stop.

The time taken is $t=15s$ .

Using the Newton's equation of linear motion,

$v=u +at$ , we find the acceleration by substituting the known values.

This implies that,

$0=37 +a(15)$

This gives us,

$0-37=15a$

$\Rightarrow -37=15a$

We divide both sides by 15 to get,

$a=-\frac{37}{15}ms^{-2}$

$a=-2.46667ms^{-2}$

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A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

2 years ago

The half-life of caffeine is 5 hours. If you ingested a 30 oz Big Gulp, how many oz of caffeine is left after one half life? * Y

xeze [42]

Answer:

The amount of caffeine left after one half life of 5 hours is 15 oz.

Explanation:

Half life is the time taken for a radioactive substance to degenerate or decay to half of its original size.

The half life of caffeine is 5 hours. So ingesting a 30 oz, this would be reduced to half of its size after the first 5 hours.

So that:

After one half life of 5 hours, the value of caffeine that would be left is;

$\frac{30}{2}$ = 15 oz

The amount of caffeine left after one half life of 5 hours is 15 oz.

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3 years ago

When the magnetic flux through a coil of wire changes, what is generated between the ends of the coil? Which equation predicts t

True [87]

Answer:

induced electromotive force (Voltage) E = - N dΦ / dt

Explanation:

When the magnetic flux this coil induces a current in each turn of the coil, which is why an induced electromotive force (Voltage) appears at the ends of the coil.

This phenomenon is fully explained by Faraday's law

E = - dΦ / dt

where in the case of a coil with N turns of has

E = - N dΦ / dt

Rl flux is the product of the normal to the area by the magnetic field, in this case the flux changes so we can assume that the area of the coil is constant

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3 years ago

A train starts at v=10m/s and accelerates at 1m/s2 for 5s. Its final speed is *

ad-work [718]

Answer:

D) 15 m/s

Explanation:

v = u + at

v = 10 m/s + 1 m/s²(5 s)

v = 15 m/s

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2 years ago

A gold sphere of radius R=100 μm and density 19g/cm^3 falls through water. Given the viscosity of water is about 10^-3 Pa s and

icang [17]

The terminal velocity of gold sphere is 39.2 cm/s

<h3>What is terminal velocity?</h3>

Terminal velocity is the maximum velocity attainable for an object as it falls through a fluid.

<h3>How to calculate the terminal velocity of the gold sphere?</h3>

The terminal velocity of the gold sphere is given by v = 2gr²(ρ - σ)/9η where

g = acceleration due to gravity = 9.8 m/s²,
r = radius of sphere = 100 μm = 100 × 10⁻⁶ m = 10⁻⁴ m = 10⁻² cm,
ρ = density of sphere = 19 g/cm³,
σ = density of water = 1.0 g/cm³ and
η = viscosity of water = 10⁻³ Pa-s

So, susbtituting the values of the variables into the equation, we have that

v = 2gr²(ρ - σ)/9η

v = 2 × 9.8m/s²× (10⁻² cm)²(19 g/cm³ - 1.0 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 9.8 m/s² × 10⁻⁴ cm² × (18 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 980 cm/s² × 10⁻⁴ cm² × 2 g/cm³/(1 × 10⁻³ Pa-s)

v = 3920 g/s² × 10⁻⁴/(1 × 10⁻³ Pa-s)

v = 392 cm/s × 10³ × 10⁻⁴

v = 392 × 10⁻¹ cm/s

v = 39.2 cm/s

So, the terminal velocity is 39.2 cm/s

Learn more about terminal velocity of sphere here:

brainly.com/question/21684177

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1 year ago