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andreev551 [17]

3 years ago

11

I’m the absence of friction,gravity and all other external forces what kind of speed will an object display if it travels from t

he bottom to the top of an incline plane

1 answer:

Natali [406]3 years ago

4 0

Velocity Im pretty sure

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A child slides down a frictionless playground slide from a height of 2.1 m above the ground. If she starts with an initial speed

baherus [9]

Answer: Final speed will be 6.59 m/s

Explanation: Initial height of slide is 2.1 m

Final height = 0.7 m

Initial speed, u = 4 m/s

Final speed, v = ??

To find the final speed in this question we can use equation of motion i.e.

$v^{2} - u^{2} = 2aS$

Here, S is the distance covered by the child on the slide i.e. net height

S = 2.1 - 0.7 = 1.4 m

and a = g, acceleration due to gravity i.e. 9.8 m/s

Put all the values in the equation,

$v^{2} = u^{2} + 2aS$

$v^{2} = 4^{2} + 2 \times 9.8 \times 1.4$

$v^{2}$ = 16 + 27.44 = 43.44

Taking square root both the sides, we get

v = 6.59 m/s

Final speed will be 6.59 m/s

8 0

3 years ago

Which of the following help the President carry out the laws of the land?

Paha777 [63]

Answer: A

Explanation:

4 0

3 years ago

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A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration

Eduardwww [97]

my answer is 1276.7 because I subtracted 88.3 miles and seconds from 1365 MI

7 0

3 years ago

In Fig., block 1 of mass m1 slides from rest along a frictionless ramp from height h = 2.50 m and then collides with stationary

Evgesh-ka [11]

Answer:

Explanation:

Given

initial height $h=2.5 m$

$m_2=2m_1$

coefficient of static friction $\mu =0.5$

When collision is elastic respective velocities after collision is

$v_1=\frac{u_1(m_1-m_2)+2m_2u_2}{m_1+m_2}$

$v_2=\frac{u_2(m_2-m_1)+2m_1u_1}{m_1+m_2}$

where $u_1$ and $u_2$ =initial velocities of object

$v_1$ and $v_2$ final velocities of object

$u_1=\sqrt{2\times 9.8\times 2.5}$

$u_1=7 m/s$

$v_2=\frac{0+2m_1\times 7}{m_1+2m_1}$

$v_2=\frac{14}{3} m/s$

using $v^2-u^2=2 as$

$0-(4.67)^2=2\times (-0.5\times 9.8)\times s$

$s=2.22\ m$

(b)Completely Inelastic

In Completely Inelastic objects stick with each other

$m_1u_1=(m_1+m_2)v$

$v=\frac{u_1}{3}=\frac{7}{3} m/s$

using $v^2-u^2=2 as$

$0-(2.33)^2=2\times (-0.5\times 9.8)\times s$

$s=0.55\ m$

7 0

4 years ago

Someone let me know if I got these 2 questions right or not pls

Alik [6]

Answer:

both are correct .... ....

5 0

3 years ago

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