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Ivahew [28]
2 years ago
5

A 46.5-kg ball has a momentum of 57.2 kg m/s. What is the ball's speed?

Physics
1 answer:
Serhud [2]2 years ago
5 0

Answer:

1.23 m/s

Explanation:

p=mv

57.2 = 46.5v

v= 57.2/46.5

v= 1.23

If you want to verify your answer, just insert the value of v in the equation.

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A racecar driver has to hold on tightly when going around a banked curve. Approximately what is the centripetal force on a 2220.
g100num [7]

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4 0
3 years ago
A heat engine performs (245 + A) J of work in each cycle while also delivering (142 + B) J of heat to the cold reservoir. Find t
Ganezh [65]

Answer:

The value is \eta  =  54.4 \%

Explanation:

From the question we are told that

    The work input is  W = ( 245 + A ) \  J

     The heat delivered is Q =  (142 + B) \  J

      The value of A is  A =  14

        The value of B  is  B  = 72

Generally the efficiency of the heat engine is mathematically represented as

          \eta  =  \frac{W}{Q_t}

Here  Q_t is the total out energy produce by the heat engine and this is mathematically represented as

           Q_t= Q + W

=>         Q_t=  245 + A + 142 + B

=>         Q_t=  390 + A+B

So

               \eta  =  \frac{245 + A }{390 + A+ B}

=>          \eta  =  0.544

=>          \eta  =  0.544 *100

=>          \eta  =  54.4 \%

5 0
3 years ago
A force of 10 N acts on an object with a mass of 10 kg. What is its acceleration? A. 1 m/s2 B. 10 m/s2 C. 2 m/s2 D. 5 m/s2
Akimi4 [234]

The acceleration exerted by the object of mass 10 kg is \mathbf{1} m / \boldsymbol{s}^{2}

Answer: Option A

<u>Explanation:</u>

According to Newton’s second law of motion, any external force acting on a body will be directly proportional to the mass of the body as well as acceleration exerted by the body. So, the net external force acting on any object will be equal to the product of mass of the object with acceleration exerted by the object. Thus,

                  Force = Mass \times Acceleration

So,

                 Acceleration=\frac{\text {Force}}{\text {Mass}}

As the force acting on the object is stated as 10 N and the mass of the object is given as 10 kg, then the acceleration will be

                  Acceleration =\frac{10 \mathrm{N}}{10 \mathrm{kg}}=1 \mathrm{m} / \mathrm{s}^{2}

So, the acceleration exerted by the object of mass 10 kg is \mathbf{1} m / \boldsymbol{s}^{2}

4 0
3 years ago
A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg
alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = -\frac{v}{u}

where

u = object distance

v = image distance

Now,

1.90 = \frac{v}{u}

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}

\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}

\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}

\frac{1}{16.5} = \frac{ 0.90}{1.90u}

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0
3 years ago
Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and th
Mariulka [41]

Answer:

a) 0.0288 grams

b) 2.6*10^{-10} J/kg

Explanation:

Given that:

A typical human  body contains about 3.0 grams of Potassium per kilogram of body mass

The abundance  for the three isotopes are:

Potassium-39, Potassium-40, and Potassium-41 with abundances are 93.26%, 0.012% and 6.728% respectively.

a)

Thus; a person with a mass of 80 kg will posses = 80 × 3 = 240 grams of potassium.

However, the amount of potassium that is present in such person is :

0.012% × 240 grams

= 0.012/100 × 240 grams

= 0.0288 grams

b)

the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is calculate as follows:

First the Dose in (Gy) = \frac{energy \ absorbed }{mass \ of \ the \ body}

= \frac{1.10*10^6*1.6*10^{-14}}{80}

= 2.2*10^{-10} \ J/kg

Effective dose (Sv) = RBE × Dose in Gy

Effective dose (Sv) =  1.2  *2.2*10^{-10} \ J/kg

Effective dose (Sv) = 2.6*10^{-10} J/kg

 

5 0
3 years ago
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