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3 years ago

Which combinations of forces would best move a large, heavy box across a room? Choose the three statements that apply.

Physics

1 answer:

sdas [7]3 years ago

6 0

The answer would probably be A two ppl push on the box from the same side

You might be interested in

What is the electric potential energy of an electron at the negative end of the cable, relative to the positive end of the cable

VashaNatasha [74]

Answer:

Electric potential energy at the negative terminal: $1.92\cdot 10^{-18}J$

Explanation:

When a particle with charge $q$ travels across a potential difference $\Delta V$ , then its change in electric potential energy is

$\Delta U = q \Delta V$

In this problem, we know that:

The particle is an electron, so its charge is

$q=-1.60\cdot 10^{-19}C$

We also know that the positive terminal is at potential

$V_+=0V$

While the negative terminal is at potential

$V_-=-12 V$

Therefore, the potential difference (final minus initial) is

$\Delta V = -12-0 = -12 V$

So, the change in potential energy of the electron is

$\Delta U = (-1.6\cdot 10^{-19})(-12)=1.92\cdot 10^{-18}J$

This means that the electron when it is at the negative terminal has $1.92\cdot 10^{-18}J$ of energy more than when it is at the positive terminal.

Since the potential at the positive terminal is 0, this means that the electric potential energy of the electron at the negative end is

$1.92\cdot 10^{-18}J$

3 0

3 years ago

If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu

Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

$\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}$ .

Volume of water:

$V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}$ .

Mass of water:

$m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}$ .

Amount of heat that the 15 L water absorbed:

$Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}$ .

What's the mass of the hot steel tool?

The specific heat of carbon steel is $0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}$ .

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

$Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}$ .

$\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}$ .

$m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}$ .

4 0

3 years ago

At the surface, atmospheric pressure is 1.013 × 10^5 Pa. People can normally snorkel down to a depth of roughly one meter. What

natulia [17]

Answer:

1.01 × 10⁵ Pa

Explanation:

At the surface, atmospheric pressure is 1.013 × 10⁵ Pa.

We need to find the total pressure on the air in the lungs of a person to a depth of 1 meter.

Pressure at a depth is given by :

$P=\rho gh$

Where

$\rho$ is the density of air, $\rho=1.225\ kg/m^3$

So,

$P=1.225\times 9.8\times 1\\\\=12\ Pa$

Total pressure, P = Atmospheric pressure + 12 Pa

= 1.013 × 10⁵ Pa + 12 Pa

= 1.01 × 10⁵ Pa

Hence, the total pressure is 1.01 × 10⁵ Pa.

5 0

3 years ago

A sled of mass m is being pulled horizontally by a constant horizontal force of magnitude F. The coefficient of kinetic friction

steposvetlana [31]

Answer:

The average velocity of the sled is vavg = s/t.

Explanation:

Hi there!

The average velocity is calculated as the traveled distance over time:

vavg = Δx/Δt

Where:

vavg = average velocity.

Δx = traveled distance.

Δt = elapsed time.

We already know the traveled distance (s) and also know the time it takes the sled to travel that distance (t). Then, the average velocity can be calculated as follows:

vavg = s/t

Have a nice day!

6 0

3 years ago

The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0

katen-ka-za [31]

Answer:

a) F = 2.66 10⁴ N, b) h = 1.55 m

Explanation:

For this fluid exercise we use that the pressure at the tap point is

Exterior

P₂ = P₀ = 1.01 105 Pa

inside

P₁ = P₀ + ρ g h

the liquid is water with a density of ρ=1000 km / m³

P₁ = 0.85 1.01 10⁵ + 1000 9.8 5

P₁ = 85850 + 49000

P₁ = 1.3485 10⁵ Pa

the net force is

ΔP = P₁- P₂

Δp = 1.3485 10⁵ - 1.01 10⁵

ΔP = 3.385 10⁴ Pa

Let's use the definition of pressure

P = Fe / A

F = P A

the area of a circle is

A = pi r² = [i d ^ 2/4

let's reduce the units to the SI system

d = 100 cm (1 m / 100 cm) = 1 m

F = 3.385 104 pi / 4 (1) ²

F = 2.66 10⁴ N

b) the height for which the pressures are in equilibrium is

P₁ = P₂

0.85 P₀ + ρ g h = P₀

h = $\frac{P_o ( 1-0.850)}{\rho \ g}$

h = $\frac{1.01 \ 10^5 ( 1 -0.85)}{1000 \ 9.8}$

h = 1.55 m

4 0

3 years ago