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Klio2033 [76]
3 years ago
11

GIZMO

Physics
1 answer:
CaHeK987 [17]3 years ago
5 0
It is A or D but I believe A
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A car has an acceleration of -5 m/s^2. Describe the car’s motion
AlladinOne [14]

Explanation:

because the acceleration is negative, this indicates a deceleration (or slowing down) . Hence we can say that:

The car is decelerating (slowing down), i.e its velocity is decreasing, at a constant rate of 5m/s².

7 0
4 years ago
Select the choice that best completes the following sentence. Simple machines A. reduce the amount of energy needed to do work.
boyakko [2]

It is actually are tools that make work easier C because i just had it on study island

7 0
3 years ago
Read 2 more answers
A person observing a flame is interested in the types of electromagnetic waves with slightly shorter and slightly longer wavelen
Nadya [2.5K]

Answer:

Ultra violet rays and infrared rays

Explanation:

The electromagnetic radiation in the order of increasing wavelength is given as

Gamma rays

X rays

Ultra violet rays

Visible radiation

Infrared rays

Microwaves

radiowaves

So, the radiations having wavelength more than the visible radiation are infrared radiations and the radiations which having the wavelength less then the visible radiation is Ultraviolet rays.

So, we observe ultra violet radiations and infrared radiations.

3 0
4 years ago
A simple pendulum has a period of 2.5 s. What is its period if its length is increased by a factor of four?
Svetach [21]

Answer:

Its period if its length is increased by a factor of four is 5 s.

Explanation:

The period of a simple pendulum is given by;

T = 2\pi \sqrt{\frac{l}{g} } \\\\\frac{T}{2\pi} = \sqrt{\frac{l}{g} } \\\\ \frac{T^2}{4\pi^2} = \frac{l}{g}\\\\\frac{T^2}{l} = \frac{4\pi^2}{g} \\\\let \ \frac{4\pi^2}{g}  \ be \ constant \\\\\frac{T_1^2}{l_1}  = \frac{T_2^2}{l_2} \\\\

Given;

initial period, T₁ = 2.5

initial length, = L₁

new length, L₂ = 4L₁

the new period, T₂ = ?

\frac{T_1^2}{l_1}  = \frac{T_2^2}{l_2} \\\\T_2^2 = \frac{T_1^2 l_2}{l_1} \\\\T_2 = \sqrt{\frac{T_1^2 l_2}{l_1}} \\\\  T_2 = \sqrt{\frac{(2.5)^2 \ \times \ 4l_1}{l_1}}\\\\  T_2 =\sqrt{(2.5)^2 \ \times \ 4}\\\\T_2 = \sqrt{25} \\\\T_2 = 5\ s

Therefore, its period if its length is increased by a factor of four is 5 s.

5 0
2 years ago
A 50 g sample of an unknown metal is heated to 90.0C. It is placed in a perfectly insulated container along with 100 g of water
Scrat [10]

Answer:

0.64 J/g°C

Explanation:

Using the formula;

Q = m × c × ∆T

Where;

Q = amount of heat

m = mass (g)

c = specific heat capacity

∆T = change in temperature (°C)

In this case:

Q (water) = - Q (metal)

mc∆T (water) = - mc∆T (metal)

According to the information in this question,

For water; m = 100g, c = 4.18J/g°C, ∆T = (25°C - 20°C)

For metal; m = 50g, c =?, ∆T = (25°C - 90°C)

mc∆T (water) = - mc∆T (metal)

100 × 4.18 × (25°C - 20°C) = - {50 × c × (25°C - 90°C)}

100 × 4.18 × 5 = - {50 × c × -65}

2090 = -{-3250c}

2090 = 3250c

c = 2090/3250

c = 0.643

c = 0.64J/g°C

7 0
3 years ago
Read 2 more answers
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