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Ber [7]
3 years ago
10

In 1970, a rocket powered car called Blue Flame achieved a maximum speed of 1.00(10 km/h (278m/s).Suppose the magnitude of the c

ar's constant acceleration is 5.56 m/s^2. If the car is initially at rest , what is the distance traveled. During its acceleration?
Physics
1 answer:
sammy [17]3 years ago
3 0

Answer:

Distance traveled during this acceleration will be 6950 m

Explanation:

Wear have given maximum speed tat will be equal to final speed of the car        v = 278 m/sec

Constant acceleration a=5.56m/sec^2

As the car starts initially starts from rest so initial velocity of the car u = 0 m/sec

From third equation of motion v^2=u^2+2as

Putting all values in equation

278^2=0^2+2\times 5.56\times s

s = 6950 m

So distance traveled during this acceleration will be 6950 m

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Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

6 0
3 years ago
A particular star cluster contains stars all with the same apparent magnitude of +4. Near the cluster (and at the same distance
777dan777 [17]

Answer:

3,000 stars.

Explanation:

Subtract the magnitude of the single star from the magnitude of the star cluster.

(+4) - (+1) = +3

Then multiply your answer by 1000

= 3,000 is a stars.

5 0
3 years ago
7. These temperatures were recorded in Pasadena for a week in April. 87 85 80 78 83 86 90 Find each of these. (a) Mean (e) Range
slega [8]

Answer:

a) Mean = 84.14

b) Median = 85

c) Mode = no mode (since there is no variable that appears more than once in this dataset)

d) Midrange = 84

e) Range = 12

f) Variance = 14.69

g) Standard deviation = 3.83

Explanation:

The raw data to be processed is

87 85 80 78 83 86 90

a) Mean = (Σx)/N

The mean is the sum of variables divided by the number of variables

x = each variable

N = number of variables = 7

Mean = (87+85+80+78+83+86+90)/7

Mean = 84.14

b) Median is the number in the middle of the dataset when the variables are arranged in ascending or descending order.

Arranging the data in ascending order

78, 80, 83, 85, 86, 87, 90

The number in the middle is the 4th number = 85

Median = 85

c) Mode is the variable that occurs the most in a distribution.

For this question, all of the variables occur only once, with no variable occurring more than once. Hence, there is no mode for this dataset.

d) Midrange is the arithmetic mean of the highest and lowest number in the dataset.

Mathematically,

Midrange = (Highest + Lowest)/2

Midrange = (90 + 78)/2

Midrange = 84

e) Range is the difference the highest and the lowest numbers in a dataset.

Range = 90 - 78 = 12

f) Variance is an average of the squared deviations from the mean.

Mathematically,

Variance = [Σ(x - xbar)²/N]

xbar = mean

Σ(x - xbar)² = (78 - 84.14)² + (80 - 84.14)² + (83 - 84.14)² + (85 - 84.14)² + (86 - 84.14)² + (87 - 84.14)² + (90 - 84.14)² = 102.8572

Variance = (102.8572)/7

Variance = 14.69

g) Standard deviation = √(variance)

Standard deviation = √(14.69)

Standard deviation = 3.83

Hope this Helps!!!

8 0
3 years ago
Read 2 more answers
Why can you plant plants, but not water water?
Grace [21]

Answer:

Because water is water and its a liquid meanwhile plants are plants living things that you can move around.

Explanation:

3 0
2 years ago
Read 2 more answers
For the strongest electromagnet, what wire gauge should be used?
alexandr1967 [171]

Answer:

Thick

Explanation:

Thicker wure enables higher AMP's, assuming your supply is capable, so indirectly thicker wire may help. Thicker wire takes up more space. So does more turns.

Hope this helped. Have a blessed day !

7 0
2 years ago
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