2 years ago

If you wanted to do calculations with speed (distance/time), which base units would you use

Physics

1 answer:

kap26 [50]2 years ago

4 0

I would use meter and second.
Hope this helps.

You might be interested in

A 55.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300. What h

MariettaO [177]

Answer:

161.86 N

Explanation:

mass of box m= 55.0 kg

weight of the box, mg= 55×9.81

g here is acceleration due to gravity =9.81 m/sec^2

coefficient of friction between the box and the surface μ= 0.3

the friction force F_s= μmg= 0.3×55×9.81

=161.86 N

to move the ball horizontal force required is 161.86 N

8 0

2 years ago

What are 3 adjectives that describe a super bowl crowd

Aleks04 [339]

Loud
Crazed
Herd-like
Irrational
Deafening
Primitive
Possessed
Tribal

5 0

2 years ago

A 438kg car is accelerating east at 2.55m/s^2. What is the total force acting east on the car

lisabon 2012 [21]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 438 × 2.55

We have the final answer as

Hope this helps you

5 0

3 years ago

Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te

seraphim [82]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy 5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant $\gamma =1.4$

specific heat is given as $=\frac{\gamma R}{\gamma -1}$

gas constant =287 J⋅kg−1⋅K−1

$Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}$

specific heat at constant volume

$Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}$

change in internal energy $= Cv(T_2 -T_1)$

$\Delta U = 717.5 (800-295) = 3.62*10^5 J kg^{-1}$

change in enthalapy $\Delta H = Cp(T_2 -T_1)$

$\Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}$

change in entropy

$\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})$

$\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})$

$\Delta S = 382.79 J kg^{-1} K^{-1}$

7 0

2 years ago

A mango is dropped and falls freely from rest. What are its position and velocity after

marissa [1.9K]

Answer:

$1s^{2}$

Explanation:

5 0

2 years ago

Read 2 more answers