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choli [55]
3 years ago
12

Which statement about the properties of matter is true? *25*

Physics
1 answer:
mojhsa [17]3 years ago
3 0

Answer:

A possible answer would be that chemical properties depend on phisical properties if and if only the phisical properties depend on the chemical ones ( see the laws of thermodinamics)

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Active listening includes all of the following EXCEPT: A. paraphrasing B. clarifying C. ignoring D. empathizing Please select th
scoray [572]

Answer:

C: Ignoring

Explanation:

On edge 2021

8 0
3 years ago
tom and ted are sitting on seprate chairs that have wheels. tom pushes ted and, in turn, he starts moving too. in which directio
IceJOKER [234]
In the direction opposite to his push.
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3 years ago
Imagine that you are going to visit your friend. Before you get there, you decide to stop at the variety store. If you walk 200
SashulF [63]

Answer:

400m

Explanation:

Brainliest? :))

Let your initial displacement from your home to the store be

Dd

>

1 and your displacement from the store to your friend’s house

be Dd

>

2.

Given: Dd

>

1 = 200 m [N]; Dd

>

2 = 600 m [S]

Required: Dd

>

T

Analysis: Dd

>

T 5 Dd

>

1 1 Dd

>

2

Solution: Figure 6 shows the given vectors, with the tip of Dd

>

1

joined to the tail of Dd

>

2. The resultant vector Dd

>

T is drawn in red,

from the tail of Dd

>

1 to the tip of Dd

>

2. The direction of Dd

>

T is [S].

Dd

>

T measures 4 cm in length in Figure 6, so using the scale of

1 cm : 100 m, the actual magnitude of Dd

>

T is 400 m.

Statement: Relative to your starting point at your home, your

total displacement is 400 m [S].

6 0
2 years ago
A hot cube of iron was heated up using 1500 J of thermal energy and was placed in a beaker of water. Before it was heated, the i
Mariana [72]

Answer:

451.13 J/kg.°C

Explanation:

Applying,

Q = cm(t₂-t₁)............... Equation 1

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

Substitute these values into equation 2

c = 1500/[0.133(45-20)]

c = 1500/(0.133×25)

c = 1500/3.325

c = 451.13 J/kg.°C

4 0
3 years ago
What must be the length of a simple pendulum if its oscillation frequency is to be equal to that of an air-track glider of mass
Anvisha [2.4K]

Answer:

the length of the simple pendulum is 0.25 m.

Explanation:

Given;

mass of the air-track glider, m = 0.25 kg

spring constant, k = 9.75 N/m

let the length of the simple pendulum = L

let the frequency of the air-track glider which is equal to frequency of simple pendulum = F

The oscillation frequency of air-track glider is calculated as;

F = \frac{1}{2\pi } \sqrt{\frac{k}{m} } \\\\F = \frac{1}{2\pi } \sqrt{\frac{9.75}{0.25} } \\\\F = 0.994 \ Hz

The frequency of the simple pendulum is given as;

F = \frac{1}{2\pi} \sqrt{\frac{g}{l} } \\\\2\pi(F) = \sqrt{\frac{g}{l} } \\\\2\pi (0.994) = \sqrt{\frac{9.8}{l} } \\\\6.2455 = \sqrt{\frac{9.8}{l} } \\\\(6.2455)2 = \frac{9.8}{l} \\\\39.006 = \frac{9.8}{l} \\\\l = \frac{9.8}{39.006} \\\\l = 0.25 \ m

Thus, the length of the simple pendulum is 0.25 m.

8 0
3 years ago
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