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3 years ago

Would someone plz help a shister outttt

Physics

1 answer:

tiny-mole [99]3 years ago

8 0

Yaaaas, do you watch James Charles!?

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What is the change in velocity of a 22-kg object that experiences a force of 15 N for

vagabundo [1.1K]

Answer:

Force = mass × acceleration

Acceleration:

${ \tt{15 = (22 \times a)}} \\ { \tt{a = \frac{15}{22} \: {ms}^{ - 2} }}$

From first Newton's equation of motion:

${ \bf{v = u + at}}$

Change = v - u:

${ \tt{v - u = (a \times t)}} \\ { \tt{v - u = ( \frac{15}{22} \times 1.2) }} \\ { \tt{v - u = 0.82 \: {ms}^{ - 2} }}$

3 0

2 years ago

Gabriel is performing an experiment in which he is measuring the energy and work being done by a ball rolling down a hill.Which

Rufina [12.5K]

I think this one's B. energy and work are both measured in joules.

7 0

2 years ago

Read 2 more answers

Which of the following has the largest m/mentum?

ASHA 777 [7]

A: The total building of Campbell high school, including the trailers and the construction area

6 0

2 years ago

A delivery truck leaves a warehouse and travels 2.60 km north. The truck makes a left turn and travels 1.25 km west before makin

yulyashka [42]

Answer:

4.19 km and 107.35 degrees north of east

Explanation:

So in the end, the truck is (2.6 + 1.4 = 4km) north and 1.25 km west from the warehouse. We can use the Pythagorean formula to calculate the magnitude and direction α of the truck displacement from the warehouse:

$s = \sqrt{s_n^2 + s_w^2} = \sqrt{4^2 + 1.25^2} = \sqrt{16 + 1.5625} = \sqrt{17.5625} = 4.19$ km

$tan\alpha = \frac{s_n}{s_w} = \frac{4}{1.25} = 3.2$

$\alpha = tan^{-1}3.2 = 1.27 rad \approx 72.65 degrees$ north or west or (180 - 72.65) = 107.35 degrees north of east

3 0

2 years ago

Please Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

arsen [322]

Answer:

Given that

speed u=4*10^6 m/s

electric field E=4*10^3 N/c

distance b/w the plates d=2 cm

basing on the concept of the electrostatices

now we find the acceleration b/w the plates to find the horizontal distance traveled by the electron when it hits the plate.

acceleration a=qE/m= $1.6*10^{-19}*4*10^3/9.1*10^{-31} =0.7*10^{15}$ = $7*10^{14}$ m/s

now we find the horizontal distance traveled by electrons hit the plates

horizontal distance

$X=u[2y/a]^{1/2}$

= $4*10^6[2*2*10^{-2}/7*10^{14}]^{1/2}$

= $3*10^{-2}$ = 3 cm

5 0

3 years ago