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sertanlavr [38]
2 years ago
8

3. A cheetah is known to be the fastest mammal on Earth, at least for short

Physics
1 answer:
Kay [80]2 years ago
4 0

Answers:

a) 198,55 s

b) 4687.98 m

Explanation:

a) Speed v is defined as the variation of the position (distance traveled d)of a body or object with time t:

v=\frac{d}{t} (1)

We are told the cheetah runs a distance of  d=5.50(10)^{2}m=550m with a speed v=10\frac{km}{h}=2.77\frac{m}{s} (Knowing 1km=1000m and 1h=3600s), and we need to find the time it takes to travel that distance.

This means we have to find t from (1):

t=\frac{d}{v} (2)

t=\frac{550m}{2.77\frac{m}{s}} (3)

t=198.55s (4)

b) Now, if we want to know the distance the cheetah will cover with v=85\frac{km}{h}=23.61\frac{m}{s} in t=198.55s, we have to isolate d from (1):

d=v.t (5)

d=(23.61\frac{m}{s})(198.55s) (6)

Finally:

d=4687.98m

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amid [387]

Answer:

13.33m/s

Explanation:

Given data

m1= 2000kg

u1= 20m/s

m2= 1500kg

u2= 0m/s

v1= 10m/s

Required

The speed of the sticks

We know that  from the expression for the conservation of momentum

m1u1+m2u2= m1v1+m2v2

2000*20+1500*0=2000*10+1500*v2

40000=20000+1500v2

collect like terms

40000-20000= 1500v2

20000= 1500v2

v2= 20000/1500

v2= 13.33 m/s

Hence the velocity of the sticks is 13.33m/s

8 0
3 years ago
(4 points) A mother with mass m1 is skating at velocity v1 behind her daughter whose mass is m2, who is skating at v2 . Instead
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Answer:

B) collision is inelastic because they stick together after collision and share a common final velocity Vf

C) M1V1 + M2V2 = (M1 + M2)Vf

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Detailed explanation and calculation is shown in the image below

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2 years ago
The turnbuckle is tightened until the tension in the cable AB equals 2.3 kN. Determine the vector expression for the tension T a
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Answer:

a) 0.83984 i + 0.41992 j - 2.0996 k KN

b) T_ac = 1.972888 KN

Explanation:

Given:

- The tension in cable AB = 2.3 KN

Find:

a) Determine the vector expression for the tension T as a force acting on member AD.

b) Also find the magnitude of the projection of T along the line AC.

Solution:

part a)

- Find unit vector AB:

                            vector (AB) = 2 i + j - 5 k

                             mag (AB) = sqrt (2^2 + 1^2 + 5^2)

                             mag (AB) = sqrt(30)

                             unit (AB) =  ( 1 / sqrt(30) )* ( 2 i + j - 5 k )

- Find Tension vector:

                             vector (T) = unit(AB)* 2.3 KN

                                              = 0.83984 i + 0.41992 j - 2.0996 k

- The projection of T onto AC can be found from the dot product of vector T to unit vector (AC)

- For unit vector (AC)

                               vector (AC) = 2 i - 2 j - 5 k

                               mag (AC) = sqrt (2^2 + 2^2 + 5^2)

                               mag (AC) = sqrt(33)

                               unit (AC) =  ( 1 / sqrt(33) )* ( 2 i - 2 j - 5 k )

- Compute the projection:

                                T_ac = vector T . unit (AC)

T_ac = (0.83984 i + 0.41992 j - 2.0996 k)  . ( 1 / sqrt(33) )* ( 2 i - 2 j - 5 k )

                                T_ac = 0.2923947572 - 0.146973786 - 1.827467232

                                T_ac = 1.972888 KN

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Answer: 26.6 J

Explanation:

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Q= c . m.  (t2 – t1)

where c= specific heat capacity (in J/gK), m= mass of the solid (in g) ,

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Replacing by the values, we get:

Q= 0.385 J/gK . 1,550 g. (77.5ºC – 33.0ºC)= 26.6 J

3 0
2 years ago
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