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Snezhnost [94]
2 years ago
3

The jamaican bobsled team hit the brakes on their sled so that it decelerates at a uniform rate of 0.43 m/s squared how long doe

s it take to stop if it travels 85 m before coming to rest
Physics
2 answers:
Digiron [165]2 years ago
8 0

The distance covered by a certain object which is travelling at a certain speed is calculated through the equation,

     d = (V₀)t + 0.5at²

where d is the distance, V₀ is the initial speed, a is deceleration, and t is the time. Substituting the known values,

    85 = (V₀)(t) + (0.5)(-0.43 m/s)(t²)

Because we are not given with the initial velocity, our answer would remain as the equation which is written above. 

Nonamiya [84]2 years ago
5 0

Answer:

Time taken, t = 19.86 seconds    

Explanation:

It is given that,

Deceleration of the sled, a = -0.43 m/s²

We have to find the time it take to stop if it travels 85 m before coming to rest i.e final velocity v = 0. Using third equation of motion as :

v^2-u^2=2as

0-u^2=2\times (-0.43\ m/s^2)\times 85\ m

u = 8.54 m/s

The initial velocity of the sled is 8.54 m/s. For finding time taken by the sled to stop can be calculated using second equation of motion as :

s=ut+\dfrac{1}{2}at^2

85=8.54t+\dfrac{1}{2}\times (-0.43)t^2

0.215t^2-8.54t+85=0

On solving the above quadratic equation we get, t = 19.86 seconds. Hence, this is the required solution.        

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Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.4 kg gibbon
Katarina [22]

Answer:

upward force acting = 261.6 N

Explanation:

given,

mass of gibbon = 9.4 kg

arm length = 0.6 m

speed of the swing

net force must provide

F_{branch} + F_{gravity}=F_{centripetal}

force of gravity = - mg

F_{branch}=F_{centripetal}-F_{gravity}

                        = \dfrac{mv^2}{r} + mg

                        = m(\dfrac{3.4^2}{0.6} +9.8)

                        =9 x 29.067

                        = 261.6 N

upward force acting = 261.6 N

7 0
2 years ago
You are piloting a helicopter which is rising vertically at a uniform velocity of 14.70 m/s. When you reach 196.00 m, you see Ba
Cloud [144]

Answer:

The ball reaches Barney  head in  t = 8 \ s

Explanation:

From the question we are told that

 The rise velocity is  v  =  14.70 \  m/s

  The height considered is h =  196 \  m

   The horizontal velocity of the large object is  v_h  =  8.50 \  m/s

   

Generally from kinematic equation  

   s = ut + \frac{1}{2} gt^2

Here s is the distance of the object from Barney head ,

        u is the velocity of the object along the vertical axis which is equal but opposite to the velocity of the helicopter

So  

     u = -14.7 m/s

So

    196  = -14.7 t  + \frac{1}{2} * 9.8 * t^2

=  4.9 t^2 - 14.7t - 196 = 0

Solving the above equation using quadratic formula  

    The value of  t obtained is  t = 8 \ s

6 0
2 years ago
Th e heat capacity of air is much smaller than that of water, and relatively modest amounts of heat are needed to change its tem
Darina [25.2K]

Answer:

Q=1005 J

t= 0.67 sec

Explanation:

Lets take condition of room is 1 atm and 25°C.

Heat capacity ,c = 21 J /K.mol

If we assume that air is ideal gas that

P V = n R T

V= 5.5\times 6.5\times 3\ m^3

V=107.25\ m^3

V=107.25\times 1000 L

V= 107250 L

At STP number of moles given as

n=\dfrac{V}{V_{at\ S.T.P.}}

V=22.4 L at S.T.P.

n=\dfrac{107250}{22.4}\ moles

n=4787.94 moles

n= 4.784 Kmoles

So heat required to raise 10°C temperature

Q = n x c x ΔT

Q = 4.78794 x 21 x 10

Q=1004.64 J

Time t

t= Q/P

P= 1.5 KW

t = 1.004.64 /1.5

t= 0.66 sec

4 0
2 years ago
Read 2 more answers
You should have observed that there are some frequencies where the output is stronger than the input. Discuss how that is even p
nydimaria [60]

Answer:

w = √ 1 / CL

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

Explanation:

This problem refers to electrical circuits, the circuits where this phenomenon occurs are series RLC circuits, where the resistor, the capacitor and the inductance are placed in series.

In these circuits the impedance is

             X = √ (R² +  (X_{C} -X_{L})² )

where Xc and XL is the capacitive and inductive impedance, respectively

            X_{C} = 1 / wC

           X_{L} = wL

From this expression we can see that for the resonance frequency

           X_{C} = X_{L}

the impedance of the circuit is minimal, therefore the current and voltage are maximum and an increase in signal intensity is observed.

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

               V = IR

Since the contribution of the two other components is canceled, this occurs for

                X_{C} = X_{L}

                1 / wC = w L

                w = √ 1 / CL

6 0
2 years ago
The force F required to compress a spring a distance x is given by F 2 F0 5 kx where k is the spring constant and F0 is the prel
IrinaVladis [17]

Answer:

a)W=8.333lbf.ft

b)W=0.0107 Btu.

Explanation:

<u>Complete question</u>

The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.

Solution

Preload = F₀=0 lbf

Spring constant k= 200 lbf/in

Initial length of spring x₁=0

Final length of spring x₂= 1 in

At any point, the force during deflection of a spring is given by;

F= F₀× kx  where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.

W=\int\limits^2_1 {} \, Fds \\\\\\W=\int\limits^2_1( {F_0+kx} \,) dx \\\\\\W=\int\limits^a_b {kx} \, dx ; F_0=0\\\\\\W=k\int\limits^2_1 {x} \, dx \\\\\\W=k*\frac{1}{2} (x_2^{2}-x_1^{2}  )\\\\\\W=200*\frac{1}{2} (1^2-0)\\\\\\W=100.lbf.in\\\\

Change to lbf.ft by dividing the value by 12 because 1ft=12 in

100/12 = 8.333 lbf.ft

work required to compress the spring, W=8.333lbf.ft

The work required to compress the spring in Btu will be;

1 Btu= 778 lbf.ft

?= 8.333 lbf.ft----------------cross multiply

(8.333*1)/ 778 =0.0107 Btu.

6 0
2 years ago
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