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2 years ago

8

Based on the data given, what is the relationship between mass, gravitational force, speed at impact, and weight of the objects?

1 answer:

Nana76 [90]2 years ago

4 0

Answer:

gfhffgfhfgf

Explanation:

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a particle is moving along a circular path having a radius of 4 in such that its position as a function of time is given by thet

ANTONII [103]

Answer:

Explanation:

Given

radius of circular path $r=4\ in.$

Position is given by

$\theta =\cos 2t---1$

Differentiate 1 to angular velocity we get

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega =-2\sin 2t----2$

Differentiate 2 to get angular acceleration

$\frac{\mathrm{d} \omega }{\mathrm{d} t}=-2^2\cos 2t ---3$

Net acceleration is the vector summation of tangential and centripetal force

$a_t=\alpha \times r$

$a_t=-4\cos 2t\times 4=-16\cos 2t$

$a_r=\omega ^2\cdot r$

$a_r=(-2\sin 2t)^2\cdot 4$

$a_r=16\sin^2(2t)$

$a_{net}=\sqrt{a_r^2+a_t^2}$

$a_{net}=\sqrt{(16\sin ^2(2t)+(-16\cos 2t)^2}$

$a_{net}=\sqrt{256\cos ^2(2t)+256\sin ^4(2t)}$

6 0

3 years ago

Calculate the acceleration of a bus that goes

Svet_ta [14]

The acceleration of the bus is 1.11 meters per second square to the direction of motion

Explanation:

Acceleration is the rate of change of velocity

The formula of the acceleration is $a=\frac{v_{2}-v_{1}}{t}$ , where

$v_{1}$ is the initial velocity

$v_{2}$ is the final velocity
t is the time

A bus that goes from 10 km/h to a speed of 50 km/h in 10 seconds

→ $v_{1}$ = 10 km/h

→ $v_{2}$ = 50 km/h

→ t = 10 seconds

Change the unite of the time from seconds to hour

→ 1 hour = 60 × 60 = 3600 seconds

→ 10 seconds = $\frac{10}{3600}=\frac{1}{360}$ hour

Substitute these values in the formula of the acceleration above

→ $a=\frac{50-10}{\frac{1}{360}}$

→ a = 14400 km/h²

To change the unit of acceleration to meter per second change the

kilometer to meter and the hour to seconds

→ 1 km = 1000 m

→ 1 hour = 3600 seconds

→ $a=\frac{14400*1000}{(3600)^{2}}=\frac{10}{9}$

→ a = 1.11 m/sec².

The acceleration of the bus is 1.11 meters per second square to the direction of motion

Learn more:

You can learn more about the acceleration in brainly.com/question/6323625

#LearnwithBrainly

4 0

3 years ago

The question is in the picture

Sedbober [7]

Answer:

e) 120m/s

Explanation:

When the ball reaches its highest point, its velocity becomes zero, meaning

$v_0-gt = 0$ .

where $v_0$ is the initial velocity.

Solving for $t$ we get

$t = \dfrac{v_0}{g}$

which is the time it takes the ball to reach the highest point.

Now, after the ball has reached its highest point, it turns around and falls downwards. After time $t_0$ since it had reached the highest point, the ball has traveled downwards and the velocity $v_f$ it has gained is

$v_f = gt_0$ ,

and we are told that this is twice the initial velocity $v_0$ ; therefore,

$v_f = 2v_0 = gt_0$

which gives

$t_0 = \dfrac{2v_0}{g}.$

Thus, the total time taken to reach velocity $2v_0$ is

$t_{tot} = t+t_0 = \dfrac{v_0}{g}+\dfrac{2v_0}{g}$

$t_{tot} = \dfrac{3v_0}{g}.$

This $t_{tot}$ , we are told, is 36 seconds; therefore,

$36= \dfrac{3v_0}{g},$

and solving for $v_0$ we get:

$v_0 = \dfrac{36g}{3}$

$v_0 = \dfrac{36s(10m/s^2)}{3}$

$\boxed{v_0 = 120m/s}$

which from the options given is choice e.

7 0

3 years ago

You lift a 10-N physics book up in the air a distance of 1.0 m, at a constant velocity

Anarel [89]

The book is lifted upward, but gravity points down, so the work done by gravity must be negative (so you can eliminate options 1 and 3).

The force exerted on the book by gravity has magnitude

<em>F</em> = <em>mg</em> = (10 N) (9.80 m/s^2) = 9.8 N ≈ 10 N

You raise the book 1.0 m in the opposite direction, so the work done is

<em>W</em> = (10 N) (-1.0 m) = -10 J

5 0

3 years ago

The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the el

Nana76 [90]

Answer:

E

Explanation:

Using Coulomb's law equation

Force of the charge = k qQ /d²

and E = F/ q

substitute for F

E = ( K Qq/ d² ) / q

q cancel q

E = KQ / d²

so twice the distance of the from the point charge will lead to the E ( electric field ) decrease by a 4 = E/4. E is inversely proportional to d²

7 0

3 years ago