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Talja [164]
3 years ago
6

What does it mean to work out your core

Physics
1 answer:
Arte-miy333 [17]3 years ago
4 0

Answer:

I'm pretty sure it means like getting abs and working out your stomach.

Explanation:

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How does the scientific process generally begin?
Murljashka [212]
There are several approaches. The most favourable one (in my opinion) is this one:
1. Asking a question
2. Doing a research (how to answer this question)
3. Creating a hypothesis (NOT a thesis!)
4. Experimenting (to prove the hypothesis)
5. Analysing results from the experiment
6. Writing a thesis
5 0
4 years ago
Read 2 more answers
A 9.0-V battery is connected to a resistor so that there is a 0.50-A current through the resistor.For how long should the batter
salantis [7]

Answer:

35.6 s

Explanation:

The power through the resistor is given by:

P=VI

where V=9.0 V is the voltage and I=0.50 A is the current. Substituting into the formula, we find

P=(9.0 V)(0.5 A)=4.5 W

The power is also equal to:

P=\frac{W}{t}

where W is the work done while t is the time taken. Since we know the work done, W=160 J, we can re-arrange the equation to find the time taken:

t=\frac{W}{P}=\frac{160 J}{4.5 W}=35.6 s

7 0
4 years ago
Jason and Guy are throwing watermelons straight up from the ground. (They’re making fruit salad.) Jason throws his watermelon at
NISA [10]

Answer:2t

Explanation:

Given

Jason throwing watermelon at speed v and it spend t time in air

time to reach max height

v_f=u_i+at'

0=v-gt'

t'=\frac{v}{g}

and time to reach bottom is also t'

t=2t'

t=2\frac{v}{g}

For guy throws his watermelon at speed 2v

So total time in air will be

t''=2\frac{2v}{g}

t''=4\frac{v}{g}

t''=2t

5 0
4 years ago
If a loaded truck that accelerates at 1 meter per second squared loses its load and has three-fourths of the origional mass, wha
Nina [5.8K]
I believe it'd accelerate at 1.25 m/s^2 instead of 1, as it lost 1/4 of its mass (.25), so now it is .25 of 1 faster.

8 0
3 years ago
A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. (a) At wha
dezoksy [38]

a) Time at which velocity is +20.0 m/s: 2.04 s

b) Time at which velocity is -20.0 m/s: 6.12 s

c) Time at which the displacement is zero: t = 0 and t = 8.16 s

d) Time at which the velocity is zero: t = 4.08 s

e) i) ii) iii) The acceleration of the boulder is always 9.8 m/s^2 downward

f) See graphs in attachment

Explanation:

a)

The motion of the boulder is a uniformly accelerated motion, with constant acceleration

a=g=-9.8 m/s^2

downward (acceleration due to gravity). So, we can use the following suvat equation:

v=u+at

where:

v is the velocity at time t

u = 40.0 m/s is the initial velocity

a=g=-9.8 m/s^2 is the acceleration

We want to find the time t at which the velocity is

v = 20.0 m/s

Therefore,

t=\frac{v-u}{a}=\frac{20-40}{-9.8}=2.04 s

b)

In this case, we want to find the time t at which the boulder is moving at 20.0 m/s downward, so when

v = -20.0 m/s

(the negative sign means downward)

We use again the suvat equation

v=u+at

And substituting

u = +40.0 m/s

a=g=-9.8 m/s^2

We find the corresponding time t:

t=\frac{v-u}{a}=\frac{-20-(+40)}{-9.8}=6.12 s

c)

To solve this part, we can use the following suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the displacement

u = +40.0 m/s is the initial velocity

a=g=-9.8 m/s^2 is the acceleration

t is the time

We want to find the time t at which the displacement is zero, so when

s = 0

SUbstituting into the equation and solving for t,

0=ut+\frac{1}{2}at^2\\t(u+\frac{1}{2}a)=0

which gives two solutions:

t = 0 (initial instant)

u+\frac{1}{2}at=0\\t=-\frac{2u}{a}=-\frac{2(40)}{-9.8}=8.16 s

which is the instant at which the boulder passes again through the initial position, but moving downward.

d)

To solve this part, we can use again the suvat equation

v=u+at

where

u = +40.0 m/s is the initial velocity

a=g=-9.8 m/s^2 is the acceleration

We want to find the time t at which the velocity is zero, so when

v = 0

Substituting and solving for t, we find:

t=\frac{v-u}{a}=\frac{0-(40)}{-9.8}=4.08 s

e)

In order to evaluate the acceleration of the boulder, let's consider the forces acting on it.

If we neglect air resistance, there is only one force acting on the boulder: the force of gravity, acting downward, with magnitude

F=mg

where m is the mass of the boulder and g the acceleration of gravity.

According to Newton's second law, the net force on the boulder is equal to the product between its mass and its acceleration:

F=ma

Combining the two equations, we get

ma=mg\\a=g

So, the acceleration of the boulder is g=9.8 m/s^2 downward at any point of the motion, no matter where the boulder is (because the force of gravity is constant during the motion).

f)

Find the three graphs in attachment:

- Position-time graph: the position of the boulder initially increases as the boulder goes upward; however, the slope of the curve decreases as the boulder goes higher (because the velocity decreases). The boulder reaches its maximum height at t = 4.08 s (when velocity is zero), then it starts going downward, until reaching its initial position at t = 8.16 s

- Velocity-time graph: the initial velocity is +40 m/s; then it decreases linearly (because the acceleration is constant), and becomes zero when t = 4.08 s. Then the velocity becomes negative (because the boulder is now moving downward) and its magnitude increases.

- Acceleration-time graph: the acceleration is constant and it is -9.8 m/s^2, so this graph is a straight horizontal line.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0
3 years ago
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