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Ket [755]
3 years ago
9

Marcus drove his Honda Prelude for 4 hours at a rate of 55 miles per hour. How far did he travel?

Physics
2 answers:
Leto [7]3 years ago
5 0
D.220 miles is the correct answer
rodikova [14]3 years ago
5 0
55 miles/hour means you travel 55 miles every hour. 

so 55*4 hours= 220 miles, so D is the answer
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Hello, I wanted an answer from a mathematician. The number 1.04 is closer to the number 1, 2, 1.25 or 1.5.
babymother [125]

Answer:

Explanation:

Of the 4 numbers given, the answer is 1 or A

If you take the absolute value of abs(1 - 1.04) you get 0.04.

(2 - 1.04) = 0.96

1.25 - 1.04 = .21

1.5 - 1.04 = 0.46

The last three are all larger than 0.04

Note: absolute value means the positive difference between 2 numbers (even  though it is negative). If it is negative, absolute value makes it positive.

3 0
2 years ago
An apparatus like the one Cavendish used to find G has large lead balls that are 5.2 kg in mass and small ones that are 0.046 kg.
Ber [7]

Answer:

The magnitude of gravitational force between two masses is 4.91\times 10^{-9}\ N.

Explanation:

Given that,

Mass of first lead ball, m_1=5.2\ kg

Mass of the other lead ball, m_2=0.046\ kg

The center of a large ball is separated by 0.057 m from the center of a small ball, r = 0.057 m

We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :

F=G\dfrac{m_1m_2}{r^2}\\\\F=6.67259\times 10^{-11}\times \dfrac{5.2\times 0.046}{(0.057)^2}\\\\F=4.91\times 10^{-9}\ N

So, the magnitude of gravitational force between two masses is 4.91\times 10^{-9}\ N. Hence, this is the required solution.

5 0
3 years ago
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is
Ronch [10]

Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s​2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is  c = 28853.78 \ m^2 /s^2

Explanation:

From the question we are told that

         The acceleration is  a =  \frac{c}{s}\   m/s^2

         The  initial position of the projectile is s= 1.5m

         The final position of the projectile is s_f =  3 \ m

          The velocity is  v = 200 \ m/s

     Generally  time  =  \frac{ds}{dv}

   and  acceleration is a =  \frac{v}{time }

so

            a = v  \frac{dv}{ds}

 =>        vdv  =  a ds

             vdv  = \frac{c}{s}  ds

integrating both sides

           \int\limits^a_b  vdv  = \int\limits^c_d \frac{c}{s}  ds

Now for the limit

          a =  200 m/s

             b = 0 m/s  

         c = s= 3 m

          d =s_f= 1.5 m

So we have  

           \int\limits^{200}_{0}  vdv  = \int\limits^{3}_{1.5} \frac{c}{s}  ds

              [\frac{v^2}{2} ] \left | 200} \atop {0}} \right.  = c [ln s]\left | 3} \atop {1.5}} \right.

            \frac{200^2}{2}  =  c ln[\frac{3}{1.5} ]

=>           c = \frac{20000}{0.69315}

              c = 28853.78 \ m^2 /s^2

     

5 0
3 years ago
An aircraft has a mass of 3.30 x 105 kg. At a certain instant during its landing, its speed is 45.2 m/s. If the braking force is
melamori03 [73]
Ultraconservative the future and I have been working with the following questions.
8 0
3 years ago
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.23 2.23 times a second. A tack is stuck in the t
elena-14-01-66 [18.8K]

Explanation:

The given data is as follows.

       Angular velocity (\omega) = 2.23 rps

     Distance from the center (R) = 0.379 m

First, we will convert revolutions per second into radian per second as follows.

             = 2.23 revolutions per second

             = 2.23 \times 2 \times 3.14 rad/s

             = 14.01 rad/s

Now, tangential speed will be calculated as follows.

  Tangential speed, v = R \times \omega

                               = 0.379 x 14.01

                               = 5.31 m/s

Thus, we can conclude that the tack's tangential speed is 5.31 m/s.

8 0
3 years ago
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