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Lelu [443]
3 years ago
12

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

of 4 m/s and the end of the slide is 0.7 m above the ground, her final speed will be: (in m/s)
Physics
1 answer:
baherus [9]3 years ago
8 0

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

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The resistance between 2 points in an electrical circuit is 1.1 Ω. What additional resistance
valentinak56 [21]

Answer:

the new resister is 11 ohms.

Explanation:

Set it  up like this.

1/x + 1/1.1 = 1                    Subtract 1/1.1 from both sides

1/x = 1 - 1/1.1

1 - 1/1.1 = 1/11

1/x = 1/11                          Cross multiply

11 = x

If 1/11 bothers you, you could do it it another way.

1 - 1/1.1 = (1.1 - 1 ) / 1.1 = 0.1 / 1.1  Multiply top and bottom by 10

0.1*10/(1.1 * 10 ) = 1 / 11

5 0
3 years ago
Rectangular plate has a voltage of +180V and plate to the first plate) has a 'voltage of -5V. Determie another rectangular plate
3241004551 [841]

For a Rectangular plate has a voltage of +180V and a 'voltage of -5V. , the second plate has the Electric field mathematically given as

E=21.5*10^3v/m

<h3>What is the field strength?</h3>

Generally, the equation for the Electric field   is mathematically given as

E=v/d

Where

v={180-(-5)}v

v=185v

Therefore

E=185/8.6*10^{-3}

E=21.5*10^3v/m

In conclusion, Electric field

E=21.5*10^3v/m

Read more about electric field

brainly.com/question/9383604

4 0
2 years ago
A boy pulls a wagon with an applied force of 40 N on frictionless surface. If the mass of the wagon is 13kg, what is the acceler
Gekata [30.6K]

Answer:

3.1 m/s²

Explanation:

Apply Newton's second law:

∑F = ma

40 N = (13 kg) a

a ≈ 3.1 m/s²

7 0
3 years ago
Are you sure? Its either tranaverse or longitudinal
4vir4ik [10]
What is either transverse or longitudinal?
 give us the rest of the question please.
5 0
3 years ago
During an auto accident, the vehicle’s air bags deploy and slow down the passengers more gently than if they had hit the windshi
Allushta [10]

Answer:

d = 0.38 m

Explanation:

As we know that the person due to the airbag action, comes to a complete stop, in 36 msec or less, and during this time, is decelerated at a constant rate of 60 g, we can find the initial velocity (when airbag starts to work), as follows:

vf = v₀ -a*t  

If vf = 0, we can solve for v₀:

v₀ = a*t = 60*9.8 m/s²*36*10⁻³s = 21.2 m/s

With the values of v₀, a and t, we can find Δx, applying any kinematic equation that relates all of some of these parameters with the displacement.

Just for simplicity, we can use the following equation:

vf^{2} -vo^{2} = 2*a*d

where vf=0, v₀ =21.2 m/s and a= -588 m/s².

Solving for  d:

d = \frac{-vo^{2}}{2*a} = \frac{(21.2m/s)^{2} }{2*588 m/s2} =0.38 m

⇒ d = 0.38 m

5 0
3 years ago
Read 2 more answers
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