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hammer [34]
3 years ago
15

Ms. Garcia’s car can go from rest to 25 m/s in 10 seconds. The car’s velocity changes at a uniform rate. How far does the car tr

avel in these 10 seconds?
Physics
1 answer:
vampirchik [111]3 years ago
4 0

We will use this equation:

s = 1/2*a*t^2 + v0*t + s0

where:

s = space traveled

a = acceleration

t = time

v0 = initial speed

s0 = initial space

In this case::

v0 = 0

s0 = 0

So our equation will look like that now:

s = 1/2 * a * t^2

let's calculate the acceleration first of all:

a = (vf - vi) / t

where vf is the final speed and vi is the initial speed. t is the time.

a = (25m/s) / 10s = 2.5 m/s^2

Now we can calculate the space:

s = 1/2 * (2.5 m/s^2) * (10s)^2 = 125m

---

Hope it was helpful! Have a great day.

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What is newton's 3rd law of physics ​
ValentinkaMS [17]

Answer:

His third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A.

4 0
3 years ago
A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnit
djverab [1.8K]

Answer:

The current is  I  = 6.68 \  A

Explanation:

From the question we are told that  

     The radius of the loop is  r =  6 \ cm  = 0.06 \ m

     The  earth's magnetic field is B_e =  0.7G=  0.7  G * \frac{1*10^{-4} T}{1 G}  = 0.7 *10^{-4} T

      The  number of turns is  N  =1

Generally the magnetic field generated by the current in the loop is mathematically represented as

        B  =  \frac{\mu_o  * N  *  I}{2 r }

Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

         B  =  B_e

=>     B_e =  \frac{\mu_o  *  N  *  I  }{ 2 * r}

     Where  \mu is the permeability of free space with value  \mu _o  =   4\pi * 10^{-7} N/A^2

       0.7  *10^{-4}=  \frac{ 4\pi * 10^{-7}  * 1 * I}{2 * 0.06}

=>     I  =  \frac{2 *  0.06 *  0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}

       I  = 6.68 \  A

3 0
3 years ago
A 2-m long string is stretched between two supports with a tension that produces a wave speed equal to vw=50.00m/s. What are the
svetoff [14.1K]

Answer

given,

Length of the string, L = 2 m

speed of the wave , v = 50 m/s

string is stretched between two string

For the waves the nodes must be between the strings

the wavelength  is given by

           \lambda = \dfrac{2L}{n}

where n is the number of antinodes; n = 1,2,3,...

the frequency expression is given by

            f = n\dfrac{v}{2L}

now, wavelength calculation

      n = 1

           \lambda_1 = \dfrac{2\times 2}{1}

                    λ₁ = 4 m

      n = 2

           \lambda_2 = \dfrac{2\times 2}{2}

                   λ₂ = 2 m

      n =3

           \lambda_3 = \dfrac{2\times 2}{3}

                    λ₃ = 1.333 m

now, frequency calculation

      n = 1

            f = n\dfrac{v}{2L}

            f_1 =1\times \dfrac{50}{2\times 2}

                    f₁ = 12.5 Hz

      n = 2

            f = n\dfrac{v}{2L}

            f_2 =2\times \dfrac{50}{2\times 2}

                    f₂= 25 Hz

      n = 3

            f = n\dfrac{v}{2L}

            f_3 =3\times \dfrac{50}{2\times 2}

                    f₃ = 37.5 Hz

8 0
3 years ago
Amanda spent 2/5 of her time after school doing homework and ¼ of her remaining time riding her bike. If she rode her bike for 4
Doss [256]

Answer:120 min

Explanation:

Given

Amanda  spent \frac{2}{5} of her time after school doing Home work

And \frac{1}{4} of her remaining  time riding her bike

It is given that she rode her bike for 45 minutes in a week

Let t be the time after school

therefore Amanda spend \frac{2t}{5} in home work and  \frac{3t}{5} time is left

From remaining \frac{3t}{5} time she spends \frac{1}{4} time riding her bike

therefore \frac{3t}{5}\times \frac{1}{4}=45

thus t=300 min

therefore time  spent on home work is \frac{2}{5}\times 300=120 min

6 0
3 years ago
In addition to an all-round white light, what light(s) must power-driven vessels less than 65.6 feet (20 meters) long exhibit wh
Lemur [1.5K]
The vessel must also have red and green side lights.

The red light is placed on the port (left) side of the boat while the green light is placed on the starboard (right) side of the vehicle. The white lights are on both the masthead (front) and stern (rear) of the boat, unless the vessel is less than 39.4 feet, in which case the front and rear white light may be combined as only one white light.
7 0
3 years ago
Read 2 more answers
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