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34kurt
3 years ago
5

A race car driver travelling at 15m*s^-1 accelerates at a constant value for 10.0s. She is now driving at a speed of 35 m*s^-1 W

hat was her acceleration? Give your answer in m*s^-2 to one significant figure . Do not include units with your answer.
Physics
1 answer:
cupoosta [38]3 years ago
6 0

The driver speeds up with acceleration <em>a</em> so that

35 m/s = 15 m/s + <em>a</em> (10.0 s)

Solve for <em>a</em> :

20 m/s = <em>a</em> (10.0 s)

<em>a</em> = (20 m/s) / (10.0 s)

<em>a</em> = 2 m/s²

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A solenoid with 3,000.0 turns is 70.0 cm long. If its self-inductance is 25.0 mH, what is its radius? (The value of μ0 is 4π x 1
nevsk [136]

Answer:

A. 2.2*10^-2m

Explanation:

Using

Area = length x L/ uo xN²

So A = 0.7m * 25 x 10^-3H /( 4π x10^-7*

3000²)

A = 17.5*10^-3/ 1.13*10^-5

= 15.5*10^-2m²

Area= π r ²

15.5E-2/3.142 = r²

2.2*10^2m

Explanation:

5 0
3 years ago
A crate is placed on an adjustable, incline board. the coefficient of static friction between the crate and the board is 0.29.
sasho [114]

Let the angle be Θ (theta)

Let the mass of the crate be m.

a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.

Normal force (N) = mg CosΘ

μ (coefficient of static friction) = 0.29

Static friction = μN = μmg CosΘ

Now, along the ramp, the equation of net force will be:

mg SinΘ - μmg CosΘ = 0

mg SinΘ = μmg CosΘ

tan Θ = μ

tan Θ = 0.29

Θ = 16.17°

b) Let the acceleration be a.

Coefficient of kinetic friction = μ = 0.26

Now, the equation of net force will be:

mg sinΘ - μ mg CosΘ = ma

a = g SinΘ - μg CosΘ

Plugging the values

a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96

a = 2.7244 - 2.44608

a = 0.278 m/s^2

Hence, the acceleration is 0.278 m/s^2

7 0
3 years ago
Luis and Aisha conducted an experiment. They exerted different forces on four objects. Their results are shown in the table.
lesya692 [45]

Answer:

Object 3 has greatest acceleration.

Explanation:

Objects               Mass                                Force

1                            10 kg                               4 N              

2                           100 grams                       20 N

3                            10 grams                         4 N

4                             1 kg                                 20 N

Acceleration of object 1,

a_1=\dfrac{F_1}{m_1}\\\\a_1=\dfrac{4}{10}\\\\a_1=0.4\ m/s^2

Acceleration of object 2,

a_2=\dfrac{F_2}{m_2}\\\\a_2=\dfrac{20}{0.1}\\\\a_2=200\ m/s^2

Acceleration of object 3,

a_3=\dfrac{F_3}{m_3}\\\\a_3=\dfrac{4}{0.01}\\\\a_3=400\ m/s^2

Acceleration of object 4,

a_4=\dfrac{F_4}{m_4}\\\\a_4=\dfrac{20}{1}\\\\a_3=20\ m/s^2

It is clear that the acceleration of object 3 is 400\ m/s^2 and it is greatest of all. So, the correct option is (3).

4 0
3 years ago
Read 2 more answers
Consider the following distribution of objects: a 2.00-kg object with its center of gravity at (0, 0) m, a 2.20-kg object at (0,
adelina 88 [10]

Answer:

body position 4 is (-1,133, -1.83)

Explanation:

The concept of center of gravity is of great importance since in this all external forces are considered applied, it is defined by

               x_cm = 1 /M   ∑ x_{i} m_{i}

               y_cm = 1 /M   ∑ y_{i} mi

Where M is the total mass of the body, mi is the mass of each element

give us the mass and position of this masses

body 1

m1 = 2.00 ka

x1 = 0 me

y1 = 0 me

body 2

m2 = 2.20 kg

x2 = 0m

y2 = 5 m

body 3

m3 = 3.4 kg

x3 = 2.00 m

y3 = 0

body 4

m4 = 6 kg

    x4=?

   y4=?

mass center position

x_cm = 0

y_cm = 0

let's apply to the equations of the initial part

X axis

    M = 2.00 + 2.20 + 3.40

    M = 7.6 kg

    0 = 1 / 7.6 (2 0 + 2.2 0 + 3.4 2 + 6 x4)

     x4 = -6.8 / 6

     x4 = -1,133 m

Axis y

    0 = 1 / 7.6 (2 0 + 2.20 5 +3.4 0 + 6 y4)

    y4 = -11/6

    y4 = -1.83 m

body position 4 is (-1,133, -1.83)

7 0
3 years ago
During a vaporization a substance changes from what to what
dusya [7]
A substance changes from liquid to gas
6 0
3 years ago
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