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Jlenok [28]
3 years ago
15

A Car travel at a speed of 200 km/hr. How far it will go in 15 mins? 27.7 km 66.1 km7.70 km50.0 km8.33 km"

Physics
1 answer:
erastovalidia [21]3 years ago
7 0

Answer:

So car will go 50 km in 15 minutes

Explanation:

We have given speed of the car = 200 km/hour

Time t = 15 minutes

We know that 1 hour = 60 minute

So 15 minute = \frac{15}{60}=0.25hour

We have to find the distance

We know that distance = speed ×time = 200×0.25=50 km

So car will go 50 km in 15 minutes

So option (d) will be the correct option

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Samples of different materials, A and B, have the same mass, but the sample
madam [21]

Answer:

<u><em>A.)</em></u> The particles that make up material A have more mass than the

particles that make up material B.

HOPE that helps!!! :)

4 0
2 years ago
Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19
Alisiya [41]

Answer:

The workdone is  W = 9.28 * 10^{3} J

Explanation:

From the question we are told that

   The height of the cylinder is  h = 0.588\ m

   The face Area is  A = 4.19 \ m^2

    The density of the cylinder is \rho  =  0.346 * \rho_w

     Where \rho_w is the density of freshwater which has a constant value

              \rho_w = 1000 kg/m^3

     

Now  

     Let the final height of the device under the water be  =  h_f

      Let  the initial volume underwater be = V_n

     Let the initial height under water be  = h_i

      Let the final volume under water be  = V_f

According to the rule of floatation

        The weight of the cylinder =  Upward thrust

This is mathematically represented as

          \rho_c g V_n = \rho_w gV_f

         \rho_c A h = \rho A h_f

So      \frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}

   =>     \frac{h_f}{h_c}  = 0.346

Now the work done is mathematically represented as  

          W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh

               =   \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.

              = \frac{g A \rho}{2}  [h^2 - h_f^2]

              = \frac{g A \rho}{2} (h^2)  [1  - \frac{h_f^2}{h^2} ]

Substituting values

        W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)

        W = 9.28 * 10^{3} J

4 0
3 years ago
Consider a river flowing toward a lake at an average velocity of 3 m/s. the river height is 90m above the lake. what is the tota
strojnjashka [21]

Kinetic energy per unit of mass is

K=\frac{v^{2} }{2}

Given, v=3m/s^{2}

Therefore,

K=\frac{(3^{2} m/s^{2} )^2}{2}

K=4.5 J/kg

Now potential energy per unit mass is

p=g\times h

Given, h=90 m

Therefore,

p= 9.8m/s^2 \times 90

p=882.9 J/kg

Thus, total mechanical energy of the river water per unit mass is

T=K+p=(4.5+882.9)J/kg

T=887.9 J/kg

OR

T=0.887 kJ/kg

6 0
3 years ago
A steady 45 N horizontal force is applied to a 15 kg object a table. The object slides against a friction force of 30 N. Calcula
lisov135 [29]

The net force on an object subject to friction is equal to the sum of the applied force and the frictional force.

Mathematically,

F_{N} = ma = F_{applied} - f_{fr}

Here, m is mass of object and a is its acceleration. We take frictional force negative because it opposes the motion of object.

Given, m=15\ kg , F_{applied} =45\ N and f_{fr} = 30\ N

Substituting these values in above formula, we get

15\ kg\times a = 45\ N -30\ N=15\ N \\\\a=\frac{15\ N}{15\ kg} =1\ m/s^2.

Thus, the acceleration of an object is 1\ m/s^2.


6 0
3 years ago
What is the fundamental frequency of a particular medium ?
VLD [36.1K]
From my research, the question has the following choices:

a.the lowest frequency at which a standing wave is possible
b. the highest frequency at which s standing wave is possiblec. the only frequency at which a standing wave is possible
d. the only frequency at which standing wave is not possible

From there, the correct answer is A.
5 0
3 years ago
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