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3 years ago

A firetruck drive 3 miles to the east and 1.2 miles to the west what is the displacement of the firetruck

Physics

1 answer:

arsen [322]3 years ago

3 0

The answer is 1.8 miles east to answer your question

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What happens to a liquid when it loses energy

NeTakaya

It can solidify, it depends on the tempeture.

8 0

3 years ago

According to the three laws of planetary motion, planetary orbits are in the shape of a/an

Talja [164]

According to the three laws of planetary motion, planetary orbits are in the shape of an "Ellipse"

In short, Your Answer would be Option B

Hope this helps!

8 0

3 years ago

What are the answers please help

aleksandr82 [10.1K]

Answer:

a.) The main scale reading is 10.2cm

b.) Division 7 = 0.07

c.) 10.27 cm

d.) 10.31 cm

e.) 10.24 cm

Explanation:

The figure depicts a vernier caliper readings

a.) The main scale reading is 10.2 cm

The reading before the vernier scale

b.) Division 7 = 0.07

the point where the main scale and vernier scale meet

c.) The observed readings is

10.2 + 0.07 = 10.27 cm

d.) If the instrument has a positive zero error of 4 division

correct reading = 10.27 + 0.04 = 10.31cm

e.) If the instrument has a negative zero error of 3 division

correct reading = 10.27 - 0.03 = 10.24cm

3 0

3 years ago

Suppose you pour 0.250 kg of 20.0°C water into a 0.600 kg aluminum pan off the stove with a temperature of 173°C. Assume that th

lapo4ka [179]

Answer:

$T_f=5.0116^{\circ}C$

Explanation:

Given:

mass of water, $m_w=0.25\ kg$

initial temperature of water, $T_i_w=20^{\circ}C$

initial temperature of pan, $T_i_p=173^{\circ}C$

mass of pan, $m_p=0.6\ kg$

mass of water evapourated, $m_v=0.03\ kg$

specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

specific heat of aluminium pan, $c_a=900\ J.kg^{-1}.K^{-1}$

latent heat of vapourization, $L=2256000\ J.kg^{-1}$

Using the equation of heat:

Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.

$m_p.c_a.(T_{ip}-T_f)=m_v.L+(m_w-m_v).c_w.(T_f-T_{iw})$

$0.6\times 900\times (173-T_f)=0.03\times 2256000+(0.25-0.03)\times 4186\times (T_f-20)$

$T_f=5.0116^{\circ}C$

5 0

3 years ago

A constant force is applied to an object, causing the object to accelerate at 10 m/s^2. What will the acceleration be if: a) The

Liula [17]

What you need to know is that the force is

F=ma

The force is the product of mass and acceleration

this means that the acceleration is

a=F/m

a) The force is halved?
this means that f will be $\frac{F}{2}$ now:

a= $\frac{F}{2m}$

So the accelaration will also he halved (it's the original acceleratation divided by 2)

b) The object's mass is halved?
a= $\frac{F}{m/2}$ =a= $\frac{F2}{m}$

which is the original acceleration times two!! so it will double

c) The force and the object's mass are both halved?
now we have

a= $\frac{F/2}{m/2}$ =a= $\frac{2F}{2m}$ =a= $\frac{F}{m}$

so they will cancel each other out and the acceleration will stay the same!

5 0

3 years ago