All categories

2 years ago

Which part of an object's rate of change best defines acceleration?

Physics

2 answers:

likoan [24]2 years ago

7 0

Answer:

Velocity

Explanation: If you are going down hill, (Which is one direction) you will gain more speed by, gravity, friction, and most importantly velocity.

grandymaker [24]2 years ago

5 0

I believe it is speed.

Hope this helps!

You might be interested in

A box is sliding up an incline that makes an angle of 14.0° with respect to the horizontal. the coefficient of kinetic friction

mezya [45]

A box is sliding up an incline that makes an angle of 14.0° with respect to the horizontal. the coefficient of kinetic friction between the box and the surface of the incline is 0.180. the initial speed of the box at the bottom of the incline is 2.20 m/s. how far does the box travel along the incline before coming to rest?

6 0

3 years ago

If i apply 280 n of force to a 40kg object, what will it's acceleration be?

nika2105 [10]

Answer:

f=ma

f=280N

m=40kg

a=?

280=40a

a=280/40

a=70N/kg

6 0

2 years ago

For thermal equilibrium at temperature Tan appropriate measure of energy is kT where k is Boltzmann's constant. Convert the foll

Schach [20]

Answer:

1 cm⁻¹ =1.44K 1 ev = 1.16 10⁴ K

Explanation:

The relationship between temperature and thermal energy is

E = K T

The relationship of the speed of light

c =λ f = f / ν 1/λ= ν

The Planck equation is

E = h f

Let's start the transformations

c = f λ = f / ν

f = c ν

E = h f

E = h c ν

E = KT

h c ν = K T

T = h c ν / K =( h c / K) ν

Let's replace the constants

h = 6.63 10⁻³⁴ J s

c = 3 10⁸ m / s

K = 1.38 10⁻²³ J / K

v = 1 cm-1 (100 cm / 1 m) = 10² m-1

T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 1 10²

A = h c / K = 1,441 10⁻²

T = 1.44K

ν = 103 cm⁻¹ = 103 10² m

T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 103 10²

T = 148K

1 Rydberg = 1.097 10 7 m

As we saw at the beginning the λ=1 / v

T = (h c / K) 1 /λ

T = 1,441 10⁻² 1 / 1,097 10⁷

T = 1.3 10⁻⁹ K

E = 1Ev (1.6 10⁻¹⁹ J /1 eV) = 1.6 10⁻¹⁹ J

E = KT

T = E/K

T = 1.6 10⁻¹⁹ /1.38 10⁻²³

T = 1.16 10⁴ K

3 0

3 years ago

A firefighter of mass 81 kg slides down a vertical pole with an acceleration of 3 m/s 2 . The acceleration of gravity is 10 m/s

natima [27]

Answer:

The force of friction that acts on him is

$F_k=567N$

Explanation:

The firefighter with an acceleration of 3m/s^2 take the gravity acceleration as 10m/s^2 isn't necessary to know the coefficient of friction just to know the force of friction:

$F=m*a$

$F=F_w-F_k$

$m*a=F_w-F_k$

$F_w=81kg*10m/s^2=810N$

Sole to Fk

$81kg*3m/s^2=810N-F_k$

$F_k=810N-243N$

$F_k=567N$

4 0

2 years ago

A child throws a baseball upward with an initial velocity of 20 m/s. The child wants to throw the baseball at least as high as t

Umnica [9.8K]

When the ball starts its motion from the ground, its potential energy is zero, so all its mechanical energy is kinetic energy of the motion:
$E= \frac{1}{2}mv^2$
where m is the ball's mass and v its initial velocity, 20 m/s.

When the ball reaches its maximum height, h, its velocity is zero, so its mechanical energy is just gravitational potential energy:
$E=mgh$

for the law of conservation of energy, the initial mechanical energy must be equal to the final mechanical energy, so we have
$\frac{1}{2}mv^2 = mgh$
From which we find the maximum height of the ball:
$h= \frac{v^2}{2g}= \frac{(20 m/s)^2}{2 \cdot 9.81 m/s^2}=20.4 m$

Therefore, the answer is yes, the ball will reach the top of the tree.

5 0

3 years ago