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3 years ago

If a 8.1g ring is heated using 10.0 ), its temperature rises 21.7°C. Calculate the specific

Physics

1 answer:

lakkis [162]3 years ago

7 0

the water specific heat will remain at 4.184.

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A 55kg cart is pushed by a force of 225 n. what is the carts acceleration?

gayaneshka [121]

F=ma so a=F/m
a=225/55=4.09 m/sec^2

3 0

4 years ago

A ball has a diameter of 3.79 cm and average density of 0.0838 g/cm3.

suter [353]

Answer: 0.258 N

Explanation:

As the density of the object is much less than the density of water, it’s clear that the buoyant force, is greater than the weight of the object, which means that in normal conditions, it would float in water.

So, in order to get the ball submerged in water, we need to add a downward force, that add to the weight, in order to compensate the buoyant force, as follows:

F = Fb – Fg

Fb= δH20* 4/3*π*(d/2)³ * g

Fg = δb* 4/3*π*(d/2)³ *g

F= (δH20- δb) * 4/3*π*(d/2)³*g

Replacing by the values of the densities, and the ball diameter, we finally get:

F= 0.258 N

3 0

4 years ago

What are the characteristics of globular cluster stars?

AlekseyPX

<span>Old age and hundreds of thousands to millions of member stars.</span>

3 0

3 years ago

Suppose you swing a ball of mass (m) in a vertical circle on a string of length (L). As you probably know from experience, there

KATRIN_1 [288]

Answer:

a) $\omega =\sqrt{\dfrac{g}{L}}$

b)N= 21.29 rpm

Explanation:

Given that

Mass of the ball =m

Length of string = L

Lets take angular speed = ω

The centripetal force on the ball

F = m ω² L

To complete the circle ,at the top condition the force due to gravity should be equal to the centripetal force

Gravity force = mg

F= mg

m ω² L = m g

ω² L = g

$\omega =\sqrt{\dfrac{g}{L}}$

When L= 2 m

Lets take g =10 m/s²

$\omega =\sqrt{\dfrac{g}{L}}$

$\omega =\sqrt{\dfrac{10}{2}}$

ω = 2.23 rad/s

To convert in rpm

$\omega =\dfrac{2\pi N}{60}$

N=Speed in rpm

$2.23 =\dfrac{2\pi N}{60}$

N= 21.29 rpm

8 0

3 years ago

A ball rolls downhill with a constant acceleration of 4m/s squared. If it started from rest,it’s velocity at the end of 3 meters

vladimir2022 [97]

Answer:

4.9 m/s

Explanation:

Since the motion of the ball is a uniformly accelerated motion (constant acceleration), we can solve the problem by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the ball in this problem,

u = 0 (it starts from rest)

$a=4 m/s^2$ is the acceleration

s = 3 m is the distance covered

Solving for v,

$v=\sqrt{u^2+2as}=\sqrt{0+2(4)(3)}=4.9 m/s$

3 0

4 years ago