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Sergeeva-Olga [200]

3 years ago

13

Using a hammer to drive a nail, it is the reaction force that brings the motion of the hammer to a stop but the _______________

that drives the nail into the wood.

1 answer:

ratelena [41]3 years ago

3 0

Action force
Hope I helped out!

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An object is moving east with a constant speed of 30 m/s for 5 seconds. What is the object's acceleration

pentagon [3]

Answer:

<h3>The answer is 6 m/s²</h3>

Explanation:

The acceleration of an object given it's velocity and time taken can be found by using the formula

$a = \frac{v}{t} \\$

where

v is the velocity

t is the time

We have

$a = \frac{30}{5} \\$

We have the final answer as

<h3>6 m/s²</h3>

Hope this helps you

5 0

3 years ago

2 H2O2-> 2H20+02

Luden [163]

Answer:

This reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen. Add 2 molecules of hydrogen peroxide and 2 molecules of water. Since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

3]Explanation: This reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen. Add 2 molecules of hydrogen peroxide and 2 molecules of water. Since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

4]Two moles of hydrogen peroxide H2O2 decomposes to produce two moles of water H2O and one mole of oxygen gas O2(g) , which then bubbles off

6 0

2 years ago

1. Driving at slower speeds than traffic flow _____________ .

r-ruslan [8.4K]

<u>The correct option is (D). The strength of electric field depends on the amount of charge that produces the field as well as the distance from the charge. </u>

<u> </u>

Further Explanation:

The electric field intensity at a point is the measure of the force exerted by a charge particle on another charge particle in the particular area of its strength.

The electric field intensity at a distance $d$ due to a static charge having charge $q$ is directly proportional to the amount of charge and inversely proportional to the square of the distance between them.

The Electric field intensity due to a charge is given as:

$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}$

Here, $E$ is the electric field intensity, $q$ is the amount of charge and $r$ is the distance of the charge from the point.

The above expression of electric field shows that the electric field intensity at a point depends on the amount of charge as well as the distance of the point from the charge.

<u>Thus, the correct option is (D). The strength of electric field depends on the amount of charge that produces the field as well as the distance from the charge. </u>

<u> </u>

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Answer Details:

Grade: Senior school

Subject: Physics

Chapter: Electrostatics

Keywords: Strength, electric field, charge, distance, electric field intensity, magnitude of charge, electrostatic, test charge, kq/r^2.

7 0

3 years ago

Read 2 more answers

Two cars are traveling along perpendicular roads, car A at 40 mi/hr, car B at 60 mi/hr. At noon, when car A reaches the intersec

Serggg [28]

Answer:

$\frac{dD}{dt} = -4 miles/hour$

negative sign indicates that the distance is decreasing with time

Explanation:

Let at any time t after noon that is 12 p.m.

distance traveled by car A = 40t

distance traveled by car B = 90-60t

then distance between the two cars at time t

$D^2= (40t)^2+(90-60t)^2$ ............1

also, at time 1 p.m.

distance $D^2= (40\times1)^2+(90-60\times1)^2$

D=50 Km

differentiating equation 1 w.r.t. t we get

$2D\frac{dD}{dt}= 2\times40t\times40+2(90-60t)(-60)$

put t= 1 and D= 50 we get

$2\times50\frac{dD}{dt}= 3200\times1-3600\times1$

$\frac{dD}{dt} = -4 miles/hour$

3 0

3 years ago

A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i

Vera_Pavlovna [14]

Answer:

Explanation:

1. We use the conservation of momentum for before the raining and after. And also we take into account that in 0.5h the accumulated water is

100kg/h*0.5h = 50kg

$p_{b}=p_{a}\\M_{b}v_{b}=(M_{b}+m_{r})v_{a}\\v_{a}=\frac{M_{b}v_{b}}{M_{b}+m_{r}}=\frac{250kg*1\frac{m}{s}}{250kg+50kg}=0.83\frac{m}{s}$

2. the momentum does not conserve because the drag force of water makes that the boat loses velocity

3. If we assume that the force of the boat before the raining is

$F=ma=m\frac{v-v_{0}}{t-t_{0}}=250kg\frac{1m}{s^{2}}=250N$

where we have assumed that the acceleration of the boat is 1m/s{2} just before the rain starts

And if we take the net force as

$F_{net}=M_{b}a_{net}=F-F_{d}=250N-bv^{2}\\F_{net}=250N-0.5N\frac{s^{2}}{m^{2}}(1\frac{m}{s})^{2}=249.5N\\a_{net}=\frac{249.5N}{M_{b}}=\frac{249.5N}{250kg}=0.99\frac{m}{s^{2}}$

where we take v=1m/s because we are taking into account tha velocity just after the rain stars.

I hope this is useful for you

regards

5 0

3 years ago