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Arisa [49]
3 years ago
11

A 73.0 kg firefighter climbs a flight of stairs 9.0 m high. how much work is required? j

Physics
1 answer:
pshichka [43]3 years ago
7 0
The strength of the fireman in vertical direction will be given by F = m * g. Then, the work done will be given by definition by W = F * d. Substituting the expression of the Force in that of the work, we have that the work will be W = m * g * d. Substituting the given values and assuming that g = 10m / s ^ 2, we have a total work of W = (73) * (10) * (9) = 6570 J
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Dovator [93]

Answer:

Explanation:

The x component is the adjacent side making up the given angle (39.4)

The vector is the hypotenuse.

The definition of the cos (x) is adjacent / hypotenuse.

cos(39.4) = adjacent / 47.3      Multiply both sides by 47.3

47.3 * cos(39.4) = adjacent      Cos(39.4) = 0.7727

adjacent = 36.55

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3 years ago
Two charges q1 and q2 exert a 90 N electrostatic force onto each other when they are 1 m apart. They are moved further away to a
Dovator [93]

Answer:

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3 years ago
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A ferry approaches shore, moving north with a speed of 6.2 m/s relative to the dock. A person on the ferry walks from one side o
jasenka [17]

Speed of Ferry is towards North with magnitude 6.2 m/s

Here if we assume that North direction is along Y axis and East is along X axis then we can say

\vec v_f = 6.2 \hat j

Now a person walk on ferry with speed 1.5 m/s towards east with respect to Ferry

so it is given as

\vec v_{pf} = 1.5 \hat i

also by the concept of relative motion we know that

\vec v_{pf} = \vec v_p - \vec v_f

now plug in all values in it

1.5 \hat i = \vec v_p - 6.2 \hat j

\vec v_p = 1.5 \hat i + 6.2 \hat j

now if we need to find the speed of the person then we need to find its magnitude

so it is given as

v = \sqrt{1.5^2 + 6.2^2}

v = 6.37 m/s

7 0
3 years ago
Converting 15 miles to kilometer
Dmitry [639]
15 miles to kilometers would be: 24.14 kilometers
8 0
3 years ago
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A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi
Dmitriy789 [7]

Answer:

a

 A =  0.081 \  m

b

The value is  u =  0.2569 \  m/s

Explanation:

From the question we are told that

   The mass is  m  =  0.750 \ kg

   The spring constant is  k  =  17.5 \  N/m

    The instantaneous speed is  v  =  39.0 \  cm/s= 0.39 \  m/s

    The position consider is  x =  0.750A  meters from equilibrium point

   

Generally from the law of  energy conservation we have that

        The kinetic energy induced by the hammer  =  The energy stored in the spring

So

          \frac{1}{2} *  m * v^2  =  \frac{1}{2}  *  k  *  A^2

Here a is the amplitude of the subsequent oscillations

=>      A =  \sqrt{\frac{m *  v^ 2 }{ k} }

=>      A =  \sqrt{\frac{0.750 *  0.39 ^ 2 }{17.5} }

=>       A =  0.081 \  m

Generally from the law of  energy conservation we have that

The kinetic energy  by the hammer  =  The energy stored in the spring at the point considered   +   The kinetic energy at the considered point

             \frac{1}{2}  * m *  v^2 = \frac{1}{2}  * k x^2 + \frac{1}{2}  * m *  u^2

=>          \frac{1}{2}  * 0.750 *  0.39^2 = \frac{1}{2}  * 17.5* 0.750(0.081 )^2 + \frac{1}{2}  * 0.750 *  u^2

=>          u =  0.2569 \  m/s

3 0
2 years ago
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