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mixas84 [53]
3 years ago
12

Daniel has a bill of 2750 on his credit card. If interest is charged at a rate of 15% p.a., calculate the amount of interest tha

t Daniel must pay for the month?
Physics
1 answer:
9966 [12]3 years ago
8 0

Answer:

412.5

Explanation:

all you have to do is multiply .15

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What is the volume of an irregularly shaped object that has a mass 3.0 grams and a density of 6.0 g/mL
pogonyaev

Answer: The volume of an irregularly shaped object is 0.50 ml

Explanation:

To calculate the volume, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of object = 6.0g/ml

mass of object = 3.0 g

Volume of object = ?

Putting in the values we get:

6.0g/ml=\frac{3.0g}{\text{Volume of substance}}

{\text{Volume of substance}}=0.50ml

Thus the volume of an irregularly shaped object is 0.50 ml

4 0
3 years ago
Two forces are applied to a 17 kg box, as shown. The box is on a smooth surface. Which statement best describes the acceleration
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3 years ago
Explain why an experiment should test only one variable at a time
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We use only one variable at a time to find the accurate result. We want to see how the result of experiment changes everytime with a single variable.
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3 years ago
A baby is dropped from a 30 m cliff on Earth. Acceleration due to gravity on Earth is about 9.8 m/s^2. Calculate the following q
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Answer:Dropping is not a good predictor of when labor will begin. In first-time mothers, dropping usually occurs 2 to 4 weeks before delivery, but it can happen earlier. In women who have already had children, the baby may not drop until labor begins.

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4 0
3 years ago
A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.
Volgvan

Answer:

A=0.199

Explanation:

We are given that  

Mass of spring=m=450 g==\frac{450}{1000}=0.45 kg

Where 1 kg=1000 g

Frequency of oscillation=\nu=1.2Hz

Total energy of the oscillation=0.51 J

We have to find the amplitude of oscillations.

Energy of oscillator=E=\frac{1}{2}m\omega^2A^2

Where \omega=2\pi\nu=Angular frequency

A=Amplitude

\pi=\frac{22}{7}

Using the formula

0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2

A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398

A=\sqrt{0.0398}=0.199

Hence, the amplitude of oscillation=A=0.199

4 0
3 years ago
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