Stars fuse most of the hydrogen in the entire star before they die. 2 poir True False

A ball is thrown up with a speed of 15m/s. How high will it go before it begins to fall? ( g = 10m/s2 )

LekaFEV [45]

Answer:

Height is 11.25m

Explanation:

Given the following data;

Initial velocity, u = 0

Final velocity, v = 15m/s

Acceleration due to gravity, g = 10m/s²

To find the height, we would use the third equation of motion;

$V^{2} = U^{2} + 2aS$

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement (height) measured in meters.

$V^{2} = U^{2} + 2aS$

Making S the subject, we have;

$S = \frac {V^{2} - U^{2}}{2a}$

But a = g = 10m/s²

Substituting into the equation, we have;

$S = \frac {15^{2} - 0^{2}}{2*10}$

$S = \frac {225 - 0}{20}$

$S = \frac {225}{20}$

S = 11.25m

Therefore, the ball will reach a height of 11.25m before it begins to fall.

4 0

2 years ago

Make the following conversion: 34.9 cL = _____ hL

aliina [53]

We know that: 1 L = 100 cL. Or 1 cL = 0.01 L. Then we will make the conversion: 34.9 cL = 34.9 / 100 L = 0.349 L. Also: 1 hL = 100 L. 0.349 L = 0.349 / 100 hL = 0.00349 hL. This can be also written as: 3.49 * 10^(-3) hL ( in the scientific notation ). Answer: 3.49 cL = 0.00349 hL

8 0

3 years ago

A snowball that will be used to build a snowman is at the top of a only hill. If the

Darina [25.2K]

Answer:

h = 24.11 m

Explanation:

Given that,

The potential energy of the snowball is 520 J

The mass of the snowball is 2.2 kg

We need to find the height of the hill. The potential energy of an object is given by the formula as follows :

$E=mgh$

g is acceleration due to gravity

h is height of the hill

$h=\dfrac{E}{mg}\\\\h=\dfrac{520}{2.2\times 9.8}\\\\h=24.11\ m$

So, the height of the hill is 24.11 m.

5 0

3 years ago

Problem 3: Thermal expansionThe steel rod has the length 2 m and cross-section area 200 cm2at the room temperature 20◦C. Weapply

zalisa [80]

Answer:

a) 2.00024 m

b) 0.036%

c) 436.67°C

Explanation:

Given

Initial length = L₀ = 2 m

Initial cross sectional Area = A₀ = 200 cm² = 0.02 m²

We can obtain initial volume = V₀ = A₀L₀ = 0.02 × 2 = 0.04 m³

Initial Temperature = T₀ = 20°C

Coefficient of linear expansivity = α = (2 × 10⁻⁶) (°C)⁻¹

a) New length of the rod after heating to 80°C

Linear expansion is given as

ΔL = L₀ × α ×ΔT

ΔL = 2 × 2 × 10⁻⁶ × (80 - 20) = 0.00024 m = 0.24 mm

New length = old length + expansion = 2 + 0.00024 = 2.00024 m

b) The percentage of the volume change of the rod.

Volume expansion is given by

ΔV = V₀ × (3α) × ΔT

Volume expansivity ≈ 3 × (linear expansivity)

ΔV = 0.04 × (3×2×10⁻⁶) × (80 - 20) = 0.0000144 m³

Percentage change in volume = 100% × (ΔV/V₀) = 100% × (0.0000144/0.04) = 0.036%

c) The maximal temperature we can allow if the volume should not increase by more than half percent.

For a half percent increase in volume, the corresponding change in volume needs to be first calculated.

Percentage change in volume = 100% × (ΔV/V₀)

0.5 = 100% × (ΔV/0.04)

(ΔV/0.04) = 0.005

ΔV = 0.0002 m³

Then we now investigate the corresponding temperature that causes this.

ΔV = V₀ × (3α) × ΔT

0.0002 = 0.04 × (3×2×10⁻⁶) × ΔT

ΔT = (0.0002)/(0.04 × 3 × 2 × 10⁻⁶) = 416.67°C

Maximal temperature = T₀ + ΔT = 20 + 416.67 = 436.67°C

4 0

2 years ago

Wich of the following celestial bodies is most likely to have many craters

Ede4ka [16]

Where are the following awnsers?

6 0

2 years ago