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Gala2k [10]
2 years ago
8

A frictionless toy car is placed on a ramp, which is inclined at an unknown angle with respect to the horizontal. Starting from

rest, the car is allowed to roll freely down the ramp. After a time interval of Δ=0.67 s, the car has traveled a distance of =1.2 m down the ramp and is moving with speed f. Find the speed of the car at the end of the 0.67 s time interval.
Physics
1 answer:
NikAS [45]2 years ago
6 0

The final speed of the toy car at the end of the given time period is 3.58 m/s.

The given parameters;

  • distance traveled by the car, s = 1.2 m
  • time of motion of the car, t = 0.67 s
  • initial velocity of the car, u = 0

The acceleration of the car is calculated as;

s = ut + \frac{1}{2} at^2\\\\1.2 = 0 + 0.5\times a\times (0.67)^2\\\\1.2 = 0.225a\\\\a = \frac{1.2}{0.225} \\\\a = 5.33 \ m/s^2

The final velocity of the toy car is calculated as;

v_f^2 = u^2 + 2as\\\\v_f^2 = 0 + 2\times 5.33 \times 1.2\\\\v_f^2 = 12.792\\\\v_f = \sqrt{12.792} \\\\v_f = 3.58 \ m/s

Thus, the final speed of the toy car at the end of the given time period is 3.58 m/s.

Learn more here: brainly.com/question/20352766

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An object was released from rest at height of 1.65 m with respect to ground. Determine the time it takes the object to reach the
vivado [14]

Answer:

The time taken by the object to reach the ground is 0.58 seconds.

Explanation:

Given that,

An object was released from rest at height of 1.65 m with respect to ground. We need to find the time taken by the object to reach the ground. Initial speed of the object is 0 as it is at rest. It will move downward under the action of gravity such that, the distance covered by the object is given by :

d=ut+\dfrac{1}{2}gt^2

d=\dfrac{1}{2}gt^2

t=\sqrt{\dfrac{2d}{g}}

t=\sqrt{\dfrac{2\times 1.65}{9.8}}

t = 0.58 seconds

So, the time taken by the object to reach the ground is 0.58 seconds. Hence, this is the required solution.

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Answer:

Explanation:

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3 years ago
The greater the mass of an object
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3 years ago
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
Elanso [62]

Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

4 0
3 years ago
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