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3 years ago

A.in which portion of the curve,

Physics

1 answer:

lozanna [386]3 years ago

8 0

Answer:

the answer is 4 acceleration is 0

Explanation:

this is because when parallel line is drawn through x axis the velocity will bw uniform

plz brainliest

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In order to make an object stop, what direction does the unbalanced force need to be applied?

nataly862011 [7]

Against the object in motion

6 0

4 years ago

Rosalinda’s take-home pay is 1700 a month she spends 11% of her take-home pay on her cable bill how much is Rosalinda’s monthly

Marizza181 [45]

Answer:

Explanation:

To solve this simply put 11% in its decimal form (0.11) and multiply it by her take-home pay and boom that's yer answer 187

6 0

3 years ago

A 3.42 kg mass hanging vertically from a spring on the Earth (where g = 9.8 m/s2) undergoes simple oscillatory motion. If the sp

sineoko [7]

Answer:

Time period of oscillation on moon will be equal to 3.347 sec

Explanation:

We have given mass which is attached to the spring m = 3.42 kg

Spring constant K = 12 N/m

We have to find the period of oscillation

Period of oscillation is equal to $T=2\pi \sqrt{\frac{m}{K}}$ , here m is mass and K is spring constant

So period of oscillation $T=2\times 3.14\times \sqrt{\frac{3.42}{12}}$

$T=2\times 3.14\times 0.533=3.347sec$

So time period of oscillation will be equal to 3.347 sec

As it is a spring mass system and from the relation we can see that time period is independent of g

So time period will be same on earth and moon

3 0

3 years ago

A cheetah is resting when a hare traveling in a straight path passes at it's top speed of 25m/s. It takes the cheetah 1.5 second

timofeeve [1]

Answer:

a) 2.3 m/s^2

b) 33.3 s

c) 803 m

d) 0 m/s

Explanation:

We need to apply the accelerated motion formulas.

The acceleration is given by:

$a=\frac{v_f-v_o}{t}\\\\a=\frac{45m/s-0m/s}{20s}\\\\a=2.3m/s^2$

we need to calculate the time the cheetah takes to stop when it starts to decelerate:

$v=vo+a*t\\\\t=\frac{v-v_o}{a}\\\\t=\frac{0-45m/s}{-4m/s^2}=11.3s\\\\t_t=20s+2s+11.3s=33.3s$

We need to calculate the displacement for all of the stages:

stage 1:

$x_1=\frac{1}{2}*a*t^2\\\\x_1=\frac{1}{2}*2.3m/s^2*(20)^2\\\\x_1=460m$

stage 2:

$x_2=v*t\\x_2=45m/s*2s=90m$

stage 3:

$x_3=v_o*t+\frac{1}{2}*a*t^2\\\\x_3=45m/s*(11.3s)-\frac{1}{2}*4m/s^2*(11.3)^2\\\\x_3=253m$

The total displacement is given by:

$d=x_1+x_2+x_3\\d=460m+90m+253m=803m$

the final velocity of the cheetah is zero because the cheetah decelerates until stop.

4 0

4 years ago

Tafu is working with subatomic particles in the physics lab. a positron is traveling in a straight line down the particle accele

KatRina [158]

Answer:

Infinity

Explanation:

When you have a limit approaching infinity one method you can use is to remember that the highest power dominants all of the other ones so when we have (t^2)/t-1 like in your problem, we see that t^2 has the highest power of 2 so if that's going to infinity and it's not being divided by any powers higher than it will just be inifinity.

4 0

3 years ago