All categories

2 years ago

Acceleration depends on which two factors?

Physics

1 answer:

Kaylis [27]2 years ago

7 0

Answer:

a = (V2 - V1) / t equation defining acceleration

Acceleration depends on velocity change and the time interval involved.

You might be interested in

In a machining operation that approximates orthogonal cutting, the cutting tool has a rake angle =100 . The chip thickness ratio

Rudik [331]

Answer:

The percentage of the total energy dissipated into shear plane is 89.46%.

Explanation:

Given that,

Rake angle = 10°

Thickness ratio= 0.5

Cutting Force = 400 N

Thrust force = 200 N

Speed =3 m/s

Suppose the shear force is 345.21 N.

We need to calculate the shear plane angle

Using formula shear angle

$\tan\phi=\dfrac{r\cos\alpha}{1-r\sin\alpha}$

Put the value in to the formula

$\tan\phi=\dfrac{0.5\cos10}{1-0.5\sin10}$

$\tan\phi=0.539$

$\phi=\tan^{-1}(0.539)$

$\phi=28.32^{\circ}$

We need to calculate the shear velocity

Using formula of shear velocity

$v_{2}=\dfrac{v\cos\alpha}{\cos(\phi-\alpha)}$

Put the value into the formula

$v_{2}=\dfrac{3\times\cos10}{\cos(28.32-10)}$

$v_{2}=3.11\ m/s$

We need to calculate the percentage of the total energy dissipated into shear plane

Using formula of energy dissipated

$\%d=\dfrac{P_{s}}{P}\times100$

$\%d=\dfrac{F_{s}\times v_{c}}{F_{c}\times v}\times100$

Put the value into the formula

$\%d=\dfrac{345.21\times3.11}{400\times3}\times100$

$d=89.46\%$

Hence, The percentage of the total energy dissipated into shear plane is 89.46%.

6 0

3 years ago

A student buys a plastic dart gun and tries to find the maximum horizontal range. The student shoots the gun straight up and it

ryzh [129]

Answer:

The value is $R_{max} = 33.54 \ m$

Explanation:

From the question we are told that

The total time of flight is $t = 3.7 \ s$

Generally from kinematic equation

$v = u - g * \frac{t}{2}$

So v is the velocity at maximum height and the value is v = 0 m/s

$0 = u - 9.8 * \frac{ 3.7}{2}$

=> $u = 18.13 \ m/s$

Here u is the initial velocity of the dart as it leaves that gun

Gnerally the horizontal range of the dart is mathematically represented as

$R = \frac{u ^2 sin 2\theta }{g}$

For maximum horizontal range the value of $\theta = 45^o$

$R_{max} = \frac{ 18.13 ^2 sin 2(45) }{9.8}$

=> $R_{max} = 33.54 \ m$

6 0

3 years ago

A 80 kg block starts from rest at the top of a 15m frictionless ramp. The block slides down the frictionless ramp onto a rough h

Nadusha1986 [10]

Answer:

W(fric) = -12000 J

Explanation:

Given that

mass ,m= 80 kg

Initial speed ,u= 0 m/s

Height ,h= 15 m

Final speed ,v= 0 m/s

We know that

Work done by all the forces = Change in the kinetic energy

$W_{gravity}+W_{fric}=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

$W_{gravity}+W_{fric}=\dfrac{1}{2}m\times 0^2-\dfrac{1}{2}m\times 0^2$

$W_{gravity}+W_{fric}=0$

W(fric)= - m g h

W(fric) = - 80 x 10 x 15 ( g=10 m/s²)

W(fric) = -12000 J

negative sign indicates that force and displacement is in opposite direction.

5 0

3 years ago

Which shows evidence of active transport?

Sunny_sXe [5.5K]

Answer: D

X and Y

Explanation:

X The least amount of magnesium Took in magnesium

Y Slightly less magnesium than the cell Took in magnesium

Because active transport occurs when ions or molecules move from less concentration region to high concentration region through semi membrane with the help of some energy.

5 0

4 years ago

Read 2 more answers

A 0.017-kg acorn falls from a position in an oak tree that is 18.5 meters above the ground.

Marianna [84]

Answer:

Part a)

Final speed of the corn is 19.05 m/s

Part b)

Kinetic energy of the corn is 3.1 J

Explanation:

Part a)

As we know that the initial position of the corn is

h = 18.5 m

now we also know that it will fall from rest and moving under constant acceleration so we will have

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(9.81)(18.5)$

$v_f = 19.05 m/s$

Part b)

Kinetic energy of the corn is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.017)(19.05)^2$

$K = 3.1 J$

4 0

4 years ago