Ask question

Ask question

All categories

3 years ago

6

Consider the equation S + O2 → SO2. What type of reaction is this? a. synthesis b. decomposition c. single displacement d. doubl

e displacement

2 answers:

arlik [135]3 years ago

7 0

Answer:

It's a synthesis reaction since S and $O_{2}$ combine to form sulphur dioxide.

Explanation:

In this reaction S and $O_{2}$ synthesized to form a single product, $SO_{2}$ ,so it will be a synthesis reaction.

VashaNatasha [74]3 years ago

4 0

This is a synthesis reaction because there are two reactants combining to form one product.

You might be interested in

mestny [16]

Answer:bsnz

Explanation:nslsosols

7 0

2 years ago

why do you believe in social learning theory and how does learning many of our morals and practice from our family uniil we grow

Hitman42 [59]

Ion know. What story is this?

4 0

2 years ago

A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it

vredina [299]

Answer:

The distance is 11 m.

Explanation:

Given that,

Friction coefficient = 0.24

Time = 3.0 s

Initial velocity = 0

We need to calculate the acceleration

Using newton's second law

$F = ma$ ...(I)

Using formula of friction force

$F= \mu m g$ ....(II)

Put the value of F in the equation (II) from equation (I)

$ma=\mu mg$ ....(III)

$a = \mu g$

Put the value in the equation (III)

$a=0.24\times9.8$

$a=2.352\ m/s^2$

We need to calculate the distance,

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}2.352\times(3.0)^2$

$s=10.584\ m\ approx\ 11\ m$

Hence, The distance is 11 m.

3 0

3 years ago

Can someone please help me with these physics problems? I just don’t even know where to start.

KIM [24]

#1

for the block of mass 5 kg normal force is given as

$F_n = mg$

$F_n = 5*9.8 = 49 N$

friction force is given as

$F_f = \mu F_n$

$F_f = 0.1*49 = 4.9 N$

Net force is given as

$F_{net} = ma$

$F_{net} = 5*2 = 10 N$

now we know that

$F_{net} = F_{app} - F_f$

$10 = F_{app} - 4.9$

$F_{app} = 14.9 N$

#2

Normal force is given as

$F_n = mg$

$F_n = 6*9.8$

$F_n = 58.8 N$

now we know that

$F_{net} = F_{app} - F_f$

$F_{net} = 0$

as object moves with constant velocity

$F_{app} = F_f = 15 N$

now for coefficient of friction we can use

$F_f = \mu F_n$

$15 = \mu * 58.8$

$\mu = 0.255$

#3

net force upwards is given as

$F = 1.2 * 10^{-4} N$

mass is given as

$m = 7 * 10^{-5} kg$

now as per newton's law we can say

$F = ma$

$1.2 * 10^{-4} = 7 * 10^{-5} * a$

$a = 1.71 m/s^2$

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

$F_{net} = F_{app} - F_f$

$ma = F_{app} - F_f$

$5*6 = 40 - F_{f}$

$F_f = 40 - 30 = 10 N$

part b)

for coefficient of friction we can use

$F_f = \mu F_n$

$10 = \mu * F_n$

here normal force is given as

$F_n = mg = 5*9.8 = 49 N$

now we have

$\mu = \frac{10}{49} = 0.204$

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

$d = v_i*t + \frac{1}{2}at^2$

$20 = 0 + \frac{1}{2}a(5^2)$

$a = 1.6 m/s^2$

now net force is given as

$F_{net} = ma$

$F_{net} = 10*1.6 = 16 N$

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

$a = \frac{v_f - v_i}{t}$

$a = \frac{0 - 25}{1.5} = -16.67 m/s^2$

now the force is gievn as

$F = ma$

$F = 5*16.67 = 83.3 N$

3 0

3 years ago

A train stops suddenly at the train station and everyone in the cars continue moving forward.

IgorC [24]

Answer:

nope

Explanation:

5 0

3 years ago