Which are functions of proteins?

What is the resultant fore acting on a 500g object accelerating 5m/s2

zimovet [89]

2.5N

Explanation:

Given parameters:

Mass of object = 500g

Acceleration = 5m/s²

unknown:

Resultant force = ?

Solution:

According to Newton's second law "the net force on a body or the resultant force is the product of its mass and acceleration".

Resultant or net force = m x a

where m is the mass

a is the acceleration

Now we need to convert the mass to kg

1000g = 1kg

500g = $\frac{500}{1000}$ = 0.5kg

Therefore;

Input the parameters:

Force = 0.5 x 5 = 2.5N

learn more:

Newton's law brainly.com/question/11411375

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3 0

3 years ago

You find it takes 200 N of horizontal force tomove an unloaded pickup truck along a level road at a speed of2.4 m/s. You then lo

IgorC [24]

ANSWER:

F(h)= 230 N is the horizontal force you will need to move the pickup along the same road at the same speed.

STEP-BY-STEP EXPLANATION:

F(h) is Horizontal Force = 200 N

V is Speed = 2.4 m/s

The total weight increase by 42%

coefficient of rolling friction decrease by 19%

Since the velocity is constant so acceleration is zero; a=0

Now the horizontal force required to move the pickup is equal to the frictional force.

F(h) = F(f)

F(h) = mg* u

m is mass

g is gravitational acceleration = 9.8 m/s^2

200 = mg*u

Since weight increases by 42% and friction coefficient decreases by 19%

New weight = 1+0.42 = 1.42 = (1.42*m*g)

New friction coefficient = μ = 1 - 0.19 = 0.81 = 0.81 u

F(h) = (0.81μ) (1.42 m g)

= (0.81) (1.42) (μ m g)

= (0.81) (1.42) (200)

= 230 N

4 0

3 years ago

A certain radioactive nuclide decays with a disintegration constant of 0.0178 h-1.

umka21 [38]

Explanation:

Given that,

The disintegration constant of the nuclide, $\lambda=0.0178\ h^{-1}$

(a) The half life of this nuclide is given by :

$t_{1/2}=\dfrac{ln(2)}{\lambda}$

$t_{1/2}=\dfrac{ln(2)}{0.0178}$

$t_{1/2}=38.94\ h$

(b) The decay equation of any radioactive nuclide is given by :

$N=N_oe^{-\lambda t}$

$\dfrac{N}{N_o}=e^{-\lambda t}$

Number of remaining sample in 4.44 half lives is :

$t_{1/2}=4.44\times 38.94$

$t_{1/2}=172.89\ h^{-1}$

So, $\dfrac{N}{N_o}=e^{-0.0178\times 172.89}$

$\dfrac{N}{N_o}=0.046$

(c) Number of remaining sample in 14.6 days is :

$t_{1/2}=14.6\times 24$

$t_{1/2}=350.4\ h^{-1}$

So, $\dfrac{N}{N_o}=e^{-0.0178\times 350.4}$

$\dfrac{N}{N_o}=0.0019$

Hence, this is the required solution.

4 0

4 years ago

Two rings of radius 5 cm are 20 cm apart and concentric with a common horizontal x-axis. The ring on the left carries a uniforml

Yanka [14]

Answer:

The electric field due to the right ring at a location midway between the two rings is $2.41\times10^{3}\ V/m$

Explanation:

Given that,

Radius of first ring = 5 cm

Radius of second ring = 20 cm

Charge on the left of the ring = +30 nC

Charge on the right of the ring = -30 nC

We need to calculate the electric field due to the right ring at a location midway between the two rings

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{qx}{(x^2+R^2)^{\frac{3}{2}}}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times30\times10^{-9}\times0.1}{((0.1)^2+(0.2)^2)^{\frac{3}{2}}}$

$E=2.41\times10^{3}\ V/m$

Hence, The electric field due to the right ring at a location midway between the two rings is $2.41\times10^{3}\ V/m$

3 0

4 years ago

Cuantos CM son:<br><br>8 newtons

lozanna [386]

Answer:

is this a question? I am confused

7 0

3 years ago