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3 years ago

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You have 2 minutes to get to PE from science class before you get a tardy. If PE is 100m away and you walk at a speed of 1.1m/s

will you get there on time?
A. No because it will take you 130 seconds to get there.

B. Yes because you will get there in 1 minute.

C. No because you should run to all of your classes.

D. Yes because it will only take 91 seconds at the speed.

2 answers:

mart [117]3 years ago

4 0

Answer:

D

Explanation:

barxatty [35]3 years ago

3 0

Answer:

its b

Explanation:

can i get brainliest

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a student is pushing a 50 kilogram cart with a force of 500 newtons another students measures the speed of the cart and finds th

Irina-Kira [14]

The force of friction is 300 N

Explanation:

We can solve the problem by applying Newton's second law of motion: in fact, the net force acting on an object is equal to the product between the mass of the object and its acceleration. So we can write

$\sum F = ma$

where

$\sum F$ is the net force acting on the object

m is its mass

a is its acceleration

For the cart in this problem, we have two forces acting on it:

- The force of push, F = 500 N, forward

- The force of friction, $F_f$ , backward

So Newton's second law can be rewritten as

$F-F_f = ma$

where

m = 50 kg

$a=4 m/s^2$ is the acceleration of the cart

And solving for $F_f$ , we find the force of friction:

$F_f = F-ma=500-(50)(4)=300 N$

Learn more about force of friction:

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3 years ago

Can some answer 7 a and b please

pav-90 [236]

Answer:

a.After $15$ second Mr Comer's speed $=$ $1.7 m/s$

b.Distance travelled by Mr.Comer in $15$ seconds $=$ $18.75\ m$

Explanation:

a. Lets recall our first equation of motion $V_{f} =V_{i} + at$

Now we know that $V_{i} = 0.8m/s$ , $a = 0.06\ m/s^2$ and

$t = 15 \ sec$

Plugging the values we have.

$V_{f} =V_{i} + at$

$V_{f} =0.8 + 0.06 \times 15$

$V_{f} =0.8+ 0.9$

$V_{f} =1.7 m/s$

Then Mr.Comer's speed after $15$ sec $=$ $1.7ms^-1$

b.

Lets find the distance and recall our third equation of motion.

$V_{f} ^2-V{i}^2 = 2as$

So $s =$ distance covered.

Dividing both sides with 2a we have.

$\frac{V_{f} ^2-V{i}^2}{2a} = 2as$

Plugging the values.

$\frac{(1.7)^2-(0.8)^2}{2\times0.06} = 2as$

$s= 18.75\ m$

So Mr.Comer will travel a distance of $s= 18.75\ m$ .

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3 years ago

A key falls from a bridge that is 45 m above the water. the key falls straight down and lands in a model boat traveling at a con

erastova [34]

Let the key is free falling, therefore from equation of motion

$h = ut +\frac{1}{2}gt^2.$ .

Take initial velocity, u=0, so

$h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2$ .

$h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2 \\\ t =\sqrt{\frac{2h}{g} }$

As velocity moves with constant velocity of 3.5 m/s, therefore we can use formula

$d= v \times t$

From above substituting t,

$d = v \times \sqrt{\frac{2h}{g} }$ .

Now substituting all the given values and g = 9.8 m/s^2, we get

$d = 3.5 \ m/s \times \sqrt{\frac{2 \times 45 m}{9.8 m/s^2} } = 10.60 m$ .

Thus, the distance the boat was from the point of impact when the key was released is 10.60 m.

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