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3 years ago

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I’m stuck in question 2. Please help

1 answer:

kicyunya [14]3 years ago

3 0

Answer:

3 parts see pic

Explanation:

see pic

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A massless rope passes over a massless pulley suspended from the ceiling. A 4-kg block is attached to one end and a 5-kg block i

borishaifa [10]

The acceleration on the 5kg block is 1.088m/s^2

Data;

mass(a) = 4kg
mass (b) = 5kg
acceleration of the 5kg block.

<h3>The Acceleration on the 5kg block</h3>

To solve this question, we just need to use Newton's second law of motion which states that "the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it."

$F_n_e_t = M_t_o_t_a_l * a\\F = ma\\F_n_e_t = mg\\$

But then the force on both sides will be

$mg = ma\\5g - 4g= (4+5) a\\(5-4)g = 9a\\g = 9a\\a = \frac{g}{9} \\a = \frac{9.80}{9} \\a = 1.088 m/s^2$

The acceleration on the 5kg block is 1.088m/s^2

Learn more on newton's second law of motion here;

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4 0

2 years ago

WILL GIVE BRAINLIEST TO THE CORRECT ANSWER

LenKa [72]

Answer:

See the answers below.

Explanation:

We can solve both problems using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F =m*a

where:

F = force [N] (units of newtons)

m = mass = 1000 [kg]

a = acceleration = 3 [m/s²]

$F = 1000*3\\F=3000[N]$

And the weight of any body can be calculated by means of the mass product by gravitational acceleration.

$W=m*g\\W=1000*9.81\\W=9810 [N]$

4 0

3 years ago

A car of mass m goes around a banked curve of radius r with speed v. If the road is frictionless due to ice, the car can still n

Sholpan [36]

Answer:

horizontal component of normal force is equal to the centripetal force on the car

Explanation:

As the car is moving with uniform speed in circle then the force required to move in the circle is towards the center of the circle

This force is due to friction force when car is moving in circle with uniform speed

Now it is given that car is moving on the ice surface such that the friction force is zero now

so here we can say that centripetal force is due to component of the normal force which is due to banked road

Now we have

$N sin\theta = \frac{mv^2}{R}$

$N cos\theta = mg$

so we have

$v = \sqrt{Rg tan\theta}$

so this is horizontal component of normal force is equal to the centripetal force on the car

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3 years ago

Who discovered that certain stars produce electromagnetic radiation, as known as pulsars

PIT_PIT [208]

Answer:

Antony Hewish

Explanation:

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3 years ago

Two particles having charges of 0.440 nC and 11.0 nC are separated by a distance of 1.80 m . 1) At what point along the line con

Harlamova29_29 [7]

Answer:

a) 0.3 m

b) r = 0.45 m

Explanation:

given,

q₁ = 0.44 n C and q₂ = 11.0 n C

assume the distance be r from q₁ where the electric field is zero.

distance of point from q₂ be equal to 1.8 -r

now,

E₁ = E₂

$\dfrac{K q_2}{(1.8-r)^2} = \dfrac{K q_1}{r^2}$

$(\dfrac{1.8-r}{r})^2= \dfrac{q_2}{q_1}$

$\dfrac{1.8-r}{r}= \sqrt{\dfrac{11}{0.44}}$

1.8 = 6 r

r = 0.3 m

<h3>b) zero when one charge is negative.</h3>

let us assume q₁ be negative so, distance from q₁ be r

from charge q₂ the distance of the point be 1.8 +r

now,

E₁ = E₂

$\dfrac{K q_2}{(1.8+r)^2} = \dfrac{K q_1}{r^2}$

$(\dfrac{1.8+r}{r})^2= \dfrac{q_2}{q_1}$

$\dfrac{1.8+r}{r}= \sqrt{\dfrac{11}{0.44}}$

1.8 =4 r

r = 0.45 m

4 0

4 years ago