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3 years ago

9

IF YOUR GOOD AT SCIENCE THEN PLEASE ANSWER THIS ASAP I WILL MARK YOU THE BRAINLIEST

1 answer:

Digiron [165]3 years ago

8 0

Answer:

This is an open circuit

Explanation:

An open circuit I believe

It needs to be closed for the bulb to be turned on

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A motorcycle rider travelling at 30m/s sees a child run into the streets 190m ahead. If the rider takes 1 second to react before

adell [148]

Answer:

The distance by which the child is safe is $k = 10 \ m$

Explanation:

From the question we are told that

The speed of the motorcycle is $v_n = 30 m/s$

The distance of the child from the motorcycle is L = 190 m

The reaction time is $t_r = 1 \ s$

deceleration is $a = -3m/s^2$

Generally the distance covered during the reaction time is

$D = v_n * t_r$

=> $D = 30 * 1$

=> $D = 30 \ m$

Generally the distance covered during the deceleration

$v^2 = u^2 + 2as$

Here v is the final velocity which is zero

So

$0 = 30 ^2 + 2 * (-3)s$

=> $s = 150 \ m$

So the total distance the motorcycle rider will cover before coming to rest is

$d = D + s$

=> $d = 30 + 150$

=> $d = 180 \ m$

Therefore the amount of distance by which the child id safe is

$k = L -d$

=> $k = 190 -180$

=> $k = 10 \ m$

5 0

3 years ago

AbbyWinterHeart, Cookiemookielookie I NEED HELP!!!

pochemuha

Answer:

the third option down is correct

4 0

3 years ago

Read 2 more answers

A skateboarder starts at 2.75 m/s and accelerates for 8.90 s over a distance of 16.7

Andrei [34K]

Answer:

1.00 m/s

Explanation:

Given:

Δx = 16.7 m

v₀ = 2.75 m/s

t = 8.90 s

Find: v

Δx = ½ (v + v₀) t

16.7 m = ½ (v + 2.75 m/s) (8.90 s)

v = 1.00 m/s

3 0

3 years ago

A friend claims that her car can accelerate from a stop to 60 mi/h (26.8 m/s) in 5.1 s , but the speedometer is broken. You deci

Lemur [1.5K]

I attached a diagram, according to the description given, to better understand the problem. The car accelerates, at a 'a' speed, which will also experience the hanging object, in two components:

The one in x, with 'a' in the opposite direction to the car's address

The one found in y, the product of gravity. The two components are related through the tangent and the respective angle, as well,

$tan\theta = \frac{ma}{mg}$

$tan\theta = \frac{a}{g}$

To identify, we know that it can be expressed as a function of speed,

$a= \frac{\Delta v}{t} = \frac{v-u_0}{5.1}\frac{26.8}{5.1}$

$a=5.26m/s^2$

Replacing in our angle formula,

$tan\theta = \frac{5.26}{9.8}$

$tan\theta = 0.54$

$\theta = tan^{-1}(0.54)$

$\theta = 28.4\°$

If there is no friction, we consider the vertical forces of both the Washes and the Normal car. So,

$F=ma$

Therefore the relationship with the vertical component would be given by,

$\Rightarrow \frac{F}{mg} = \frac{ma}{mg} = \frac{a}{g}=\frac{5.26}{9.8}=0.54 =54\%$

4 0

3 years ago

Which of the following is an oxidation-reduction reaction? ZnS(s) + 2O2(g) mc011-1.jpg ZnSO4(s) CaO(s) + H2O(l) mc011-2.jpg Ca(O

aliya0001 [1]

<span>ZnS(s) + 2O2(g) mc011-1.jpg ZnSO4(s) CaO(s) + H2O(l) </span>

4 0

4 years ago

Read 2 more answers