A car is moving at 35 mph and comes to a stop in 5 seconds.

Find the acceleration of the car.

Answer:

The spring constant is 222.2 N/m

Explanation:

Given;

mass of the spring, m = 200 g = 0.2 kg

extension of the spring, x = 15 cm = 0.15 m

speed of the mass, v = 5 m/s

spring constant, k = ?

Apply law of conservation of energy;

¹/₂kx² = ¹/₂mv²

k = mv² / x²

k = (0.2 x 5²) / (0.15²)

k = 222.2 N/m

Therefore, the spring constant is 222.2 N/m

That's a good, simple description of a "hypothesis".

Another way to describe it is "an educated guess".

Once you have it in words, it's time to start checking it out, with experiments that can show whether it's true or not.  

If your experiments seem to show that your hypothesis seems to be true, that doesn't 'prove' it, but you can start calling your hypothesis a "theory".  

(It's possible that you may never be able to 'prove' it.  It may remain a theory forever. Like gravity, germs, atoms, and relativity.  Thousands of successful experiments don't 'prove' a theory, but it can be trashed by one good, valid experiment to show that it's false.)

Answer:

I think the answer for this question is number b. gravitational

Answer:

Below

Explanation:

s = Distance

v = Velocity ( speed)

t = time

[tex]s = vt[/tex]

Divide both sides of the equation by t

[tex] \frac{s}{t} = \frac{vt}{t} \\ \frac{s}{t} = v[/tex]

Answer:

3.72 m/s

Explanation:

Answer:

It will be Less than 100J

Explanation:

Because the second ball has both kinetic and potential energy so just before it hits the ground the kinetic energy will be total energy-potential energywhich will be less than 100J

Answer:

-6 m/s²

Explanation:

Given:

v₀ = 50 m/s

v = 20 m/s

t = 5 s

Find: a

v = at + v₀

20 m/s = a (5 s) + 50 m/s

a = -6 m/s²

Answer:

-5

Explanation:

Answer:

The value is  [tex]a =  -3.7 \  m/s^2 [/tex]

Explanation:

From the question we are told that

   The  landing speed is  [tex]u =  71.9 \  m/s[/tex]

   The  distance traveled is  [tex]d =  675 \  m[/tex]

    The velocity it is reduced to is  [tex]v  =  11.3 \  m/s[/tex]

Generally the average acceleration is mathematically represented as

      [tex]a =  \frac{ v^2  -  u^2 }{ 2 * d }[/tex]

=>  [tex]a =  \frac{ 11.3^2 - 71.9^2 }{ 2 * 675 }[/tex]

=>   [tex]a =  -3.7 \  m/s^2 [/tex]

Answer:

C) increases from left to right across the periodic table

Explanation:

if you're reading this, you better smile cause you aren't the only one stressed out about school :))) have a great day love

The trend of the reactivity of non metals in the periodic table is that it increases from left to right across the periodic table.

The periodic table is composed of metals on the left hand side and nonmetals on the right hand side.

As such, nonmetallic property increases from left to right across a period and so does the reactivity of nonmetals.

This implies that as we move from left to right across a period, nonmetallic elements become more reactive.

Learn more: https://brainly.com/question/23040395

Answer:

Total current, [tex]I=4.51\times 10^{-14}\ A[/tex]

Explanation:

Velocity of electrons is 7130 m/s and particles are evenly spaced with 0.0288 m between electrons. We can find no of electrons passing per second as follows :

[tex]n_e=\dfrac{7130\ m/s}{0.0288\ m}\\\\n_e=247569.44[/tex]

Velocity of protons is 2583 m/s and particles are evenly spaced with 0.0747 m between electrons. We can find no of protons passing per second as follows :

[tex]n_p=\dfrac{2583 \ m/s}{0.0747 \ m}\\\\n_p=34578.31[/tex]

Total current in this region is equal to sum of current due to electrons and current due to protons.

[tex]I=n_e\times e+n_p\times e\\\\I=e(n_e+n_p)\\\\I=1.6\times 10^{-19}\times (247569.44+34578.31)\\\\I=4.51\times 10^{-14}\ A[/tex]

Hence, this is the required solution.

Answer:

d) 4

Explanation:

The image attached shows the options.

The resultant vector is the resultant of two or more vectors. The resultant vector is gotten by adding the sum of the displacement of the vectors together (that is the sum of all the individual vectors).

From the question, since runner A runs north and runner B runs east the resultant vector would be the sum of the displacement of vectors A and B. Also, the direction of the vector would be the north east starting from the beginning of vector A to end of vector B. The correct option is d) 4

Answer:

no its not like the undertow in the ocean

Explanation:

Answer:

Yes

Explanation:

This is possible due to Bernoulli's principle.

Answer:

The body must be moving with a constant non zero acceleration.

Explanation:

Force produces acceleration on any mass it is applied on. The acceleration produced depends on the magnitude and direction of the force. For this block being dragged by a constant horizontal force, The body will undergo a constant non-zero acceleration that will steadily increase its velocity along the direction of the force.

Answer:

Explanation:

a

The angle each thread is making with the vertex = 16.5 degree

weight acting downwards = 66.7 x 10⁻³ x 9.8 N .

distance between two charges = 2 x 84.8 x 10⁻² sin 33/2

= .48 m

repulsive force between the two charges

= 9 x 10⁹ x q² / .48²

F = 39 x 10⁹ q²

Now, the tension in the thread be T

T cos 16.5 = mg

T sin 16.5 = F

dividing

tan 16.5 = F / mg = 39 x 10⁹ x q² / 66.7 x 10⁻³ x 9.8

.296 x  66.7 x 10⁻³ x 9.8 = 39 x 10⁹ x q²

q² = 4.96 x 10⁻¹²

q = 2.23 x 10⁻⁶ C

= 2.23 μC

Answer:

17.115m/s

Explanation:

The formula that will be used to calculate the speed of the truck is expressed as shown:

v = √ar

a is the acceleration of the truck

r is the radius of curvature of the road at the top of the road plus centre of mass G who can be considered a particle, is 2 m

v = √a(R+G)

Given parameters

a = 0.3g (g is the acceleration due to gravity)

R = 98m

G = 2m

Substituting the values into the formula:

v = √0.3(9.81){98+2}

v = √0.3×9.81×100

v = √294.3

v = 17.155m/s

Hence the speed of the truck is 17.155m/s

Answer:constructive interfrence

Explanation:

Answer:

The value is [tex]n = 1.527 *10^{14} \ electrons [/tex]

Explanation:

From the question we are told that

The mass of the sphere is m = 18.5 kg

The diameter is [tex]d= 25.0 \ cm = 0.25 \ m[/tex]

The number of electrons is [tex]N = 8.10 *10^{15} \ electrons[/tex]

The mass of the plastic ball is [tex]m_b = 0.120 g[/tex]

The initial acceleration of the ball is [tex]u = 1525\ m/s^2[/tex]

Generally the radius of the sphere is mathematically evaluated as

[tex]r = \frac{d}{2}[/tex]

=> [tex]r = \frac{0.25}{2}[/tex]

=> [tex]r = 0.125 \ m [/tex]

Generally the force between the ball and the sphere is

[tex]F = \frac{k * q_1 * q_2 }{r^2}[/tex]

Generally this force can also be mathematically represented as

[tex]F = m_b * a[/tex]

So

[tex] m_b * a = \frac{k * q_1 * q_2 }{r^2}[/tex]

Here [tex]q_1[/tex] is the charge removed from the sphere whicn is evaluated as

[tex]q_1 = N * e[/tex]

substituting [tex]1.60*10^{-19} \ C[/tex] for e

[tex]q_1 = 8.10 *10^{15} * 1.60*10^{-19}[/tex]

[tex]q_1 = 0.0013 \ C [/tex]

and [tex]q_2[/tex] is the charge to be removed from the ball

So

[tex] 0.120 * 1525 = \frac{9*10^9 * 0.0013* q_2 }{0.125^2}[/tex]

[tex]q_2 = 2.44 *10^{-7} \ C [/tex]

Generally the number of electron to be removed from the ball is mathematically represented as

[tex]n = \frac{q_2}{e}[/tex]

=>    [tex]n  =  \frac{ 2.44 *10^{-7}}{1.60*10^{-19}}[/tex]

=> [tex]n = 1.527 *10^{14} \ electrons [/tex]

Answer:

The value  is   [tex]  C_u   =   5 *10^{-6} F[/tex]

Explanation:

From the question we are told that

   The voltage of the capacitor is  [tex]V  =  90 \  V[/tex]

   The  capacitance of the capacitor is  [tex]C  =  30 \ \mu F =  30 *10^{-6} \  F[/tex]

   The  final voltage is  [tex]V_a =  20 \  V[/tex]

Since the capacitors are connected in parallel we have that

      [tex]Q_u  =  Q'_u  +  Q[/tex]

Where  [tex]Q_u[/tex] is the charge of the known capacitor before it is connected to the known capacitor

     [tex]Q_u' [/tex] is the charge of the known capacitor after it is connected

       [tex]Q  [/tex] is the charge of the unknown capacitor

Also the potential across the two capacitors will be the same (except for a loss due heat (it been converted to heat ))

So

            [tex]CV  =  CV_a +  C_u *  V_a[/tex]

=>          [tex]  C_u *  V_a   =  CV - CV_a [/tex]

=>          [tex]  C_u   =   \frac{CV - CV_a }{ V_a } [/tex]

=>          [tex]  C_u   =   \frac{ [30*10^{-6} *90] - [30*10^{-6} * 20] }{ 20 } [/tex]

=>        [tex]  C_u   =   5 *10^{-6} F[/tex]

Answer:

a

   [tex]v  =  5.39 \  m/s[/tex]

b

Horizontal component

     [tex]v_x  =  5.00 \  m/s[/tex]

vertical component

     [tex]v_y  =  - 2.0 \  m/s[/tex]

c

     [tex]t =  0.921 \  s[/tex]

d

[tex]d = 4.605 \  m [/tex]

Explanation:

Generally from the question we can deduce that he initial velocity of the cork, as seen by an observer on the ground in terms of the x  unit vector is  

     [tex]v_x  =  5.00 \  m/s[/tex] due to the fact that the cork is moving horizontally

Generally from the question we can deduce that the vertical and horizontal  components of the initial velocity is  

       [tex]v_y  =  - 2.0 \  m/s[/tex] due to the fact that the balloon is moving downward which is the negative which will also cause the cork to move vertically with the balloon speed

Generally the  initial velocity (magnitude and direction) of the cork, as seen by an observer on the ground is mathematically represented as

       [tex]v  =  \sqrt{ v^2 _x  + v^2 _y  }[/tex]

       [tex]v  =  \sqrt{ 5^2  + (-2)^2 _y  }[/tex]

        [tex]v  =  5.39 \  m/s[/tex]

Generally the  initial direction of motion as seen by the same observer is mathematically represented as

    [tex]\theta =  tan^{-1}[\frac{2}{5} ][/tex]

    [tex]\theta  =  21.80^o[/tex]

Generally the time taken by the cork in the air before landing is mathematically represented as

       [tex]D  =  ut  + \frac{1}{2} g t^2[/tex]

So  D =  6 \  m from the question

     g =  9.8 \  m/s^2

     u  =  [tex]v_y[/tex] =  2 m/s  this because we are considering the  vertical motion

So

     [tex]6  =   2 t  + \frac{1}{2} *  9.8*  t^2[/tex]

       [tex]6  =   2 t  + 4.9  t^2[/tex]

Solving using quadratic formula w have that

      [tex]t =  0.921 \  s[/tex]

Generally the distance of the cork from the balloon is mathematically represented as

     [tex]d = v_x  *  t[/tex]

    [tex]d = 5  * 0.921 [/tex]

      [tex]d = 4.605 \  m [/tex]

Answer:

Below, the definition including its position of Roman people is clarified.

Explanation:

Hello there! :)

[tex]\huge\boxed{\text{C. 293K}}[/tex]

Room temperature is approximately 20°C.

We can automatically eliminate choices B and D since they are not equal to 20°C.

Since some choices use the Kelvin scale, we can convert from Celsius to Kelvin using a simple formula:

K = C° + 273

Find room temperature in degrees Kelvin:

K = 20° + 273

K = 293°

Thus, the correct choice would be C. 293K.

Explanation:

760 mmHg (millimeters of mercury) is a measure of atmospheric pressure.  It represents the height of a column of mercury at which the static pressure at the bottom is equal to the atmospheric pressure.

1 atm = 760 mmHg = 101,325 Pa = 14.7 psi

Answer:

2.98 second

Explanation:

The severity index is defined by :

[tex]S=a^{5/2}t[/tex]

a is dimensionless constant that equals the number of multiples of g

Conditions are given as :

Initial velocity, u = 0

Acceleration, a = 34 m/s²

Final velocity, v = 16.4 km/h = 4.56 m/s

We can find t from the above data as follows :

[tex]t=\dfrac{v-u}{a}\\\\t=\dfrac{4.56-0}{34}\\\\t=0.134\ s[/tex]

As a is the acceleration that is multiple of g.

So,

[tex]a=\dfrac{34}{9.8}=3.46[/tex]

So,

Severity index,

[tex]S=a^{5/2}t\\\\S=(3.46)^{5/2}\times 0.134\\\\S=2.98\ s[/tex]

Hence, the severity index for the collision is 2.98 seconds.

Explanation:

7.9x10^9 km is equal to

=7900000000km

Please mark it as brainliast

Answer:

The requested distance is 4320 meters

Explanation:

We can use the formula for velocity in this movement at constant velocity (v), which is defined as the quotient between the distance covered divided the time it took:

[tex]v=\frac{distance}{time}[/tex]

Since we know the velocity and the time, we can solve for the distance:

[tex]\neq v=\frac{distance}{time} \\18\,\frac{m}{s} = \frac{distance}{240\,\,s} \\distance=18\,*\,240\,\,m\\distance= 4320\,\,m[/tex]