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3 years ago

Which of the following is a unit of acceleration?

Physics

1 answer:

olasank [31]3 years ago

7 0

${\tt{\red{\underline{\underline{\huge{Answer:}}}}}}$

$\longrightarrow$ The rate of change of velocity per unit time is called acceleration.

$\longrightarrow$ Its SI unit is m/s².

$\huge\boxed{\fcolorbox{blue}{red}{Thank you}}$

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Can anyone plz give me a right answer now?

grandymaker [24]

Answer:

Dy = - 0.0789 [m]

Explanation:

The vertical component of the vector can be determined with the sine of the angle.

Dy = 0.250*sin(18.4)

Dy = 0.0789 [m]

As the y-component is pointing downwards the component is negative.

Dy = - 0.0789 [m]

4 0

3 years ago

A 900 kg car is traveling at 20 m/s along the road. What force must be applied to the car to stop it in a distance of 30 m2 Assu

serg [7]

Answer:

A. 6000 N

Explanation:

v²=u²+2as

0²=20²+2x30xa

-400=60a

a=-400/60

a =-6.667m/s²

f =ma

f = 900 x 6.667 = 6003N

F = 6000N

5 0

3 years ago

What are the factors that affect the weather?

jarptica [38.1K]

Answer:

temperature, atmosphere pressure,wind, topography, humidity etc

6 0

3 years ago

A non-conducting sphere of radius R = 3.0 cm carries a charge Q = 2.0 mC distributed uniformly throughout its volume. At what di

BlackZzzverrR [31]

Answer:

$r =3 *\sqrt{2} = 4.24 cm$

Explanation:

given data

Radius of sphere 3.0 cm

charge Q = 2.0 m C

We know that maximum electric field is given as

$E_{MAX}= \frac{KQ}{r^{2}}$

electric field inside the sphere can be determine by using below relation

$\frac{KQ}{r^{2}}= \frac{1}{2}*\frac{KQ}{R^{2}}$

$r = \sqrt{2}R$

$r =3 *\sqrt{2} = 4.24 cm$

4 0

3 years ago

A 50.0 kg crate is pulled 375 N of force applied to a rope. The crate slides without friction.

LUCKY_DIMON [66]

Hi there!

We can use the work-energy theorem to solve.

Recall that:

$\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}$

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

$W = \frac{1}{2}(50)(5.61^2) = 786.8025 J$

Now, we can define work:

$\large\boxed{W = Fdcos\theta}}$

Now, plug in the values:

$786.8025 = Fdcos\theta\\\\786.8025 = (375)(3.07)cos\theta$

Solve for theta:

$cos\theta = .6834\\\theta = cos^{-1}(.6834) = \boxed{46.887^o}$

4 0

3 years ago