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tino4ka555 [31]
3 years ago
9

X = v xo t. x = (10.0 m/s)(3.53 s) x = ????

Physics
2 answers:
LenaWriter [7]3 years ago
7 0

x = V<em>x</em> * t

given V<em>x</em> = 10m/s n t = 3.53s

x = 10 * 3.53

= 35.3m


labwork [276]3 years ago
7 0

The equation given is confusing: x = v xo t. But given x = (10.0 m/s)(3.53 s), I think x = (10.0 m/s)(3.53 s) = 35.3m


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A 7-kg bowling ball moving at 4 m/s strikes a 1 kg bowling pin. If the ball slows to 2 m/s in 0.05 s, how much force does it exe
iren [92.7K]

Answer:

280 N

Explanation:

acceleration = v2-v1 / time taken = (2-4 )/ 0.05 = -40 m/s^2  ( neg sign indicates slowing down )

force exerted = ma = 7 kg x -40 m/s^2 = - 280 N ( neg sign means opposite direction of initial velocity )

since the 7 kg ball is slowing down, the  direction of force will be opposite of the initial velocity , and it will be 280 N

7 0
3 years ago
At takeoff, a commercial jet has a speed of 72 m/s. Its tires have a diameter of 0.89 m. Part (a) At how many rev/min are the ti
34kurt

Answer:

a) Revolutions per minute = 2.33

b) Centripetal acceleration = 11649.44 m/s²

Explanation:

a) Angular velocity is the ratio of linear velocity and radius.

Here linear velocity = 72 m/s

Radius, r  = 0.89 x 0. 5 = 0.445 m

Angular velocity

         \omega =\frac{72}{0.445}=161.8rad/s

Frequency

         f=\frac{2\pi}{\omega}=\frac{2\times \pi}{161.8}=0.0388rev/s=2.33rev/min

Revolutions per minute = 2.33

b) Centripetal acceleration

               a=\frac{v^2}{r}

  Here linear velocity = 72 m/s

  Radius, r  = 0.445 m

Substituting

   a=\frac{72^2}{0.445}=11649.44m/s^2

Centripetal acceleration = 11649.44m/s²

6 0
3 years ago
The power of the kettle was 2.6 kW
faltersainse [42]

Answer:

Cp = 4756 [J/kg*°C]

Explanation:

In order to calculate the specific heat of water, we must use the equation of energy for heat or heat transfer equation.

Q = m*Cp*(T_f - T_i)/t

where:

Q = heat transfer = 2.6 [kW] = 2600[W]

m = mass of the water = 0.8 [kg]

Cp = specific heat of water [J/kg*°C]

T_f  = final temperature of the water = 100 [°C]

T_i = initial temperature of the water = 18 [°C]

t = time = 120 [s]

Now clearing the Cp, we have:

Cp = Q*t/(m*(T_f - T_i))

Now replacing

Cp = (2600*120)/(0.8*(100-18))

Cp = 4756 [J/kg*°C]

7 0
3 years ago
15. यदि दिइएको केन्द्र 0 भएको वृत्तको
miv72 [106K]

Answer:

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Explanation:

i don't understand the languge u used please can you change it

8 0
2 years ago
tyhe mass of an object is 117 g adding 1200j of heat will raise the temperture of the object by 12 celsius what is the specifc h
Sati [7]

Answer:

C 0.85 j/g*k

Explanation:

The specific heat capacity of a material is given by:

C_s = \frac{Q}{m \Delta T}

where

Q is the amount of heat supplied to the object

m is the mass of the object

\Delta T is the increase in temperature of the object

For the object in this problem, we have

m = 117 g is the mass

Q = 1200 J is the heat supplied

\Delta T=12^{\circ} is the increase in temperature

Substituting into the formula, we find the specific heat:

C_s = \frac{1200 J}{(117 g)(12^{\circ})}=0.85 J/gC

8 0
3 years ago
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