All categories

3 years ago

The rate at which a candle burns in millimeters per minute is:

1 answer:

3 0

since both components, length and time, are measurable
<span>since Rate = length ÷ time </span>
<span>∴ rate is also measurable and ∴ quantitative.

</span>

You might be interested in

Find the tension in the two groups that are holding the 2.9 kg object in Pl., One makes an angle of 35.6° with respect to the ve

Stella [2.4K]

Answer:

≈ 20.35 N [newton's of tension]

Explanation:

( (2.9 × 9.8) ÷ cos(35.6°) ) × sin (35.6°) =

( (28.42) ÷ (≈0.813) ) × (≈0.582) =

(≈34.96) × (≈0.582) = 20.3449446.... ≈ 20.35

4 0

2 years ago

A horizontal uniform bar of mass 2.5 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to t

11Alexandr11 [23.1K]

Answer: I am not sure if you wanted me to answer this or not.

Explanation:

6 0

2 years ago

Calculate the wavelength of an orange light wave with a frequency of 5.085 x 10^14 Hz. The speed of light is 3.0 x 10^8 m/s.

Margarita [4]

Answer:

5.9 x 10⁻⁷m

Explanation:

Given parameters:

Frequency = 5.085 x 10¹⁴Hz

Speed of light = 3.0 x 10⁸m/s

Unknown:

Wavelength of the orange light = ?

Solution:

The wavelength can be derived using the expression below;

wavelength = $\frac{v}{f}$

v is the speed of light

f is the frequency

wavelength = $\frac{3 x 10^{8} }{5.085 x 10^{14} }$ = 5.9 x 10⁻⁷m

3 0

3 years ago

A car and a lorry are about to collide. When they collide the two vehicles become tightly locked together. The lorry is going at

BartSMP [9]

Answer:

The speed of the vehicles immediately after the collision is 5.84 m/s.

Explanation:

The speed of the vehicles after the collision can be found by conservation of linear momentum:

$p_{i} = p_{f}$

$m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

Where:

m₁: is the mass of the car = 0.5 ton = 500 kg

m₂: is the mass of the lorry = 9.5 ton = 9500 kg

$v_{1_{i}}$ : is the initial speed of the car = 40 km/h = 11.11 m/s

$v_{2_{i}}$ : is the initial speed of the lorry = 20 km/h = 5.56 m/s

$v_{1_{f}}$ : is the final speed of the car =?

$v_{2_{f}}$ : is the final speed of the lorry =?

Since the two vehicles become tightly locked together after the collision $v_{1_{f}}$ = $v_{2_{f}}$ :

$m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = v(m_{1} + m_{2})$

$v = \frac{m_{1}v_{1_{i}} + m_{2}v_{2_{i}}}{m_{1} + m_{2}} = \frac{500 kg*11.11 m/s + 9500 kg*5.56 m/s}{500 kg + 9500 kg} = 5.84 m/s$

Therefore, the speed of the vehicles immediately after the collision is 5.84 m/s.

I hope it helps you!

8 0

2 years ago

A.it is moving to the right

Bas_tet [7]

Answer:

I think c buti don't think I'm correct

7 0

2 years ago