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2 years ago

Which units are used to measure force? newtons feet miles grams

Physics

2 answers:

Free_Kalibri [48]2 years ago

5 0

Newton is used to measure force

The most used unit for the force measurement is "newton" which is the SI unit of the force

DedPeter [7]2 years ago

4 0

Answer:

newtons

Explanation:

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A student drops two metallic objects into a 120-g steel con- tainer holding 150 g of water at 25°C. One object is a 200-g cube o

marissa [1.9K]

Answer:

The mass of the aluminum chunk is 258 g

Explanation:

Given;

mass of steel container = 120-g

mass of water = 150 g

initial temperature of water, = 25°C

mass of copper cube, $M_{cu}$ = 200 g

initial temperature of the copper cube, $T_c_u$ = 85°C

initial temperature of the aluminum chunk $T_A_l$ = 5.0°C

Neglecting heat loss, heat exchanged by the two metallic objects is the same since initial temperature is equal to final temperature of water.

$M_{Al}C_{Al} \delta T_{Al} = M_{cu}C_{cu} \delta T_{cu}$

where;

$C_{AL}$ is specific heat capacity of aluminum

$\delta T_{Al}$ is change in temperature of aluminum

$C_c_u$ is the specific heat capacity of copper

$\delta T_c_u$ is the change in temperature of copper

$M_{Al}C_{Al} \delta T_{Al} = M_{cu}C_{cu} \delta T_{cu} \\\\M_{Al} = \frac{M_{cu}C_{cu} \delta T_{cu}}{C_{Al} \delta T_{Al}} \\\\M_{Al} = \frac{0.2*387*60}{900*20} = 0.258 \ kg$

Therefore, the mass of the aluminum chunk is 258 g

7 0

3 years ago

A diver 40 m deep in 10 degrees C fresh water exhales a 1.5 cm diameter bubble.

zzz [600]

Answer:

0.0257259766982 m

Explanation:

$P_2$ = Atmospheric pressure = 101325 Pa

$d_1$ = Initial diameter = 1.5 cm

$d_2$ = Final diameter

$\rho$ = Density of water = 1000 kg/m³

h = Depth = 40 m

The pressure is

$P_1=P_2+\rho gh\\\Rightarrow P_1=101325+1000\times 9.81\times 40\\\Rightarrow P_1=493725\ Pa$

From ideal gas law we have

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow \dfrac{P_1\dfrac{4}{3\times8}\pi d_1^3}{T_1}=\dfrac{P_2\dfrac{4}{3\times8}\pi d_2^3}{T_2}\\\Rightarrow \dfrac{P_1d_1^3}{T_1}=\dfrac{P_2d_2^3}{T_2}\\\Rightarrow d_2=(\dfrac{P_1d_1^3T_2}{P_2T_1})^{\dfrac{1}{3}}\\\Rightarrow d_2=(\dfrac{493725\times 0.015^3\times (20+273.15)}{101325\times (10+273.15)})^{\dfrac{1}{3}}\\\Rightarrow d_2=0.0257259766982\ m$

The diameter of the bubble is 0.0257259766982 m

8 0

2 years ago

Choose all the answers that apply. Which of the following statements are true? Science is always a positive force. Animal testin

Ilya [14]

Skepticism is a quality that scientists need. Because if they do not have that, they would not be motivated to do research on things they have doubts on. We would not have things we have today. Please rate me, and if this helped, thank me. If this really helped crown me brainiest answer.

4 0

2 years ago

An airplane traveling 919 m above the ocean at 505 m/s is to drop a case of twinkies to the victims below. How much time before

Elena L [17]

Answer:

a. 13.7 s b. 6913.5 m

Explanation:

a. How much time before being directly overhead should the box be dropped?

Since the box falls under gravity we use the equation

y = ut - 1/2gt² where y = height of plane above ocean = 919 m, u = initial vertical velocity of airplane = 0 m/s, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the airplane to be directly overhead.

So,

y = ut - 1/2gt²

y = 0 × t - 1/2gt²

y = 0 - 1/2gt²

y = - 1/2gt²

t² = -2y/g

t = √(-2y/g)

So, t = √(-2 × 919 m/-9.8 m/s²)

t = √(-1838 m/-9.8 m/s²)

t = √(187.551 m²/s²)

t = 13.69 s

t ≅ 13.7 s

So, the box should be dropped 13.69 s before being directly overhead.

b. What is the horizontal distance between the plane and the victims when the box is dropped?

The horizontal distance x between plane and victims, x = speed of plane × time it takes for box to drop = 505 m/s × 13.69 s = 6913.45 m ≅ 6913.5 m

6 0

3 years ago

I need help with questions b and d, that’s all. Thank you.

Mkey [24]

b). The power depends on the RATE at which work is done.

Power = (Work or Energy) / (time)

So to calculate it, you have to know how much work is done AND how much time that takes.

In part (a), you calculated the amount of work it takes to lift the car from the ground to Point-A. But the question doesn't tell us anywhere how much time that takes. So there's NO WAY to calculate the power needed to do it.

The more power is used, the faster the car is lifted. The less power is used, the slower the car creeps up the first hill. If the people in the car have a lot of time to sit and wait, the car can be dragged from the ground up to Point-A with a very very very small power ... you could do it with a hamster on a treadmill. That would just take a long time, but it could be done if the power is small enough.

Without knowing the time, we can't calculate the power.

...

d). Kinetic energy = (1/2) · (mass) · (speed squared)

On the way up, the car stops when it reaches point-A.

On the way down, the car leaves point-A from "rest".

WHILE it's at point-A, it has no speed. So it has no (zero) kinetic energy.

7 0

2 years ago