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3 years ago

Questions

Physics

1 answer:

astra-53 [7]3 years ago

8 0

Answer:

1968

Explanation:

2400*20.5*0.004

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A boy pushes a 50 kg wheelbarrow with a velocity of 0.5 m/s. If he uses a force of 10 N, how far does he push

labwork [276]

Answer:

d=1.25 m

Explanation:

fd=.5mv2-.5mv1; 10(d)=.5(50)(.5)-0; 10(d)=12.5; d=12.5/10; d=1.25 m

4 0

3 years ago

Object A has a mass of 8.0 kg and is accelerating at 4.0 m/s2. Object B has a mass of 10.0 kg and is accelerating at 3.0 m/s2. O

Leviafan [203]

Answer:

b. B, A, C

Explanation:

To solve this problem, let us find the net forces they are experiencing.

Net force = mass x acceleration

Object A; mass = 8kg and acceleration = 4m/s²

Net force = 8 x 4 = 32N

Object B; mass = 10kg and acceleration = 3m/s²

Net force = 10 x 3 = 30N

Object C: mass = 7kg and acceleration = 5m/s²

Net force = 7 x 5 = 35N

So, increasing order of their net force;

B < A < C

6 0

3 years ago

A car is 200 m from a stop sign and traveling toward the sign at 40.0 m/s. At this time, the driver suddenly realizes that she m

victus00 [196]

Answer:

The acceleration of the car will be $a=9600m/sec^$

Explanation:

We have given that distance from stop sign s = 200 m

Time t = 0.2 sec

We have to find the constant acceleration

Now from second equation of motion $s=ut+\frac{1}{2}at^2$

$200=40\times 0.2+\frac{1}{2}\times a\times 0.2^2$

$a=9600m/sec^$

So the acceleration of the car will be $a=9600m/sec^$

6 0

4 years ago

A particle travels in a circular orbit of radius 21 m. Its speed is changing at a rate of 23.1 m/s2 at an instant when its speed

Yuri [45]

The particle has an acceleration vector with one component directed toward the center of its orbit, and the other directing tangentially to its orbit. Call these components $\vec a_c$ ( $c$ for center) and $\vec a_t$ ( $t$ for tangent). Then its acceleration vector has magnitude

$|\vec a|=\sqrt{\|\vec a_c\|^2+\|\vec a_t\|^2}$

We have

$\|\vec a_c\|=\dfrac{\|\vec v\|^2}r$

where $\|\vec v\|$ is the particle's speed and $r$ is the radius of orbit, so

$\|\vec a_c\|=\dfrac{\left(37.2\frac{\rm m}{\rm s}\right)^2}{21\,\rm m}=65.9\dfrac{\rm m}{\mathrm s^2}$

We're given that the particle's speed changes at a rate of 23.1 m/s^2. Its velocity vector points in the same direction as $\vec a_t$ , i.e. perpendicular to $\vec a_c$ , so

$\|\vec a_t\|=23.1\dfrac{\rm m}{\mathrm s^2}$

Then the magnitude of the particle's acceleration is

$\|\vec a\|=\sqrt{\left(65.9\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(23.1\dfrac{\rm m}{\mathrm s^2}\right)^2}=\boxed{69.8\dfrac{\rm m}{\mathrm s^2}}$

7 0

3 years ago

How much work is done if you push a box 200 meters with a force of 35 newtons answer?

VashaNatasha [74]

Work = force × distance
= 35 N × 200 m
= 7000 J

4 0

3 years ago