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3 years ago

What formula allows you to calculate the x component of a projectile?

Physics

2 answers:

Kay [80]3 years ago

8 0

X-component of a projectile in flight =

(initial x-component)

plus

(initial horizontal component of velocity) x (flight time so far)

RideAnS [48]3 years ago

4 0

Answer:

Explanation:

X- component = Horizontal component

Initial velocity = U

X- component =Ucos x angle

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Learning Task 2: Prepare a basin with half-filled water and stone. Drop a stone

ELEN [110]

Answer:

1 . What happens when you drop the stone?

Depending on the weight from which the stone was dropped, the glass might well break

2 depending on the size and weight and shape on the stone the glass might well break

3 depending on the density on the stone the stone might when float on the water

Explanition :

GIVE ME BRAINLESS PLEASE !!

6 0

3 years ago

A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500

jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5 ×10 ⁻³ kg

initial velocity of stone ( $u_{stone}$ ) = 0 m/s

Initial velocity of bullet ( $u_{bullet}$ ) = (500 m/s)i

Speed of the bullet after collision ( $v_{bullet}$ ) = (300 m/s) j

Suppose we represent $(v_{stone})$ to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

$m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}$

$(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}$

$(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j= (0.100 \ kg)v_{stone}$

$v_{stone}= (1.25\ kg.m/s)i-(0.75\ kg m/s)j$

$v_{stone}= (12.5\ m/s)i-(7.5\ m/s)j$

∴

The magnitude now is:

$v_{stone}=\sqrt{ (12.5\ m/s)^2-(7.5\ m/s)^2}$

$\mathbf{v_{stone}= 14.6 \ m/s}$

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

$\theta = tan^{-1} (\dfrac{-7.5}{12.5})$

$\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}$

5 0

3 years ago

When a scientist contrasts two or more objects what is he or she looking for?

tekilochka [14]

When your Contrasting multiple objects, your looking for any differences that may or may not be their. Think Compare and Contrast if that helps, The answer to this is A

5 0

3 years ago

Read 2 more answers

Many caterpillars construct cocoons from silk, one of the strongest naturally occurring materials known. Each thread is typicall

joja [24]

Answer:

a) N = 145,833,674.52174 = 1.458 × 10⁸ strands

b) diameter of single rope with the same effect = 2.415 cm

Explanation:

Hooke's law explains that stress is directly proportional to strain.

Stress ∝ Strain.

Stress = E × Strain

E = constant of proportionality = Young's Modulus = 4.0 ✕ 10⁹ N/m².

Stress = (Load/Total Cross sectional Area)

Load = a pair of 85 kg mountain climbers = 85 × 2 × 9.8 = 1666 N

Total Cross sectional Area = (Number of strands) × (Area of one strand) = A

Strain = (ΔL/L)

ΔL = 1.00 cm = 0.01 m

L = 11 m

Strain = (0.01/11) = 0.0009091

Stress = (Young's Modulus) × (Strain) = (4.0 ✕ 10⁹) × (0.0009091) = 3,636,363.64 N/m²

(Load/ total Area) = 3,636,363.64

Total area = (Load/3,636,363.64) = (1666/3,636,363.64) = 0.00045815 m²

Recall,

Total Cross sectional Area = (Number of strands) × (Area of one strand)

Area of one strand = (πd²/4)

diameter of one strand = 2 μm = (2×10⁻⁶) m

Area of one strand = (πd²/4)

= π × (2×10⁻⁶)² ÷ 4 = (3.142 × 10⁻¹²) m²

Total Cross sectional Area = (Number of strands) × (Area of one strand)

0.00045815 = N × (3.142 × 10⁻¹²)

N = 145,833,674.52174 = 1.458 × 10⁸ strands

b) If it was a single rope, the cross sectional Area would just be equal to the total cross sectional Area obtained in (a)

A = 0.00045815 m²

A = (πD²/4)

where D = diameter of the single rope

0.00045815 = (πD²/4)

D² = (4×0.00045815) ÷ π = 0.0005833347

D = 0.02415 m = 2.415 cm

Hope this Helps!!!

6 0

4 years ago

Read 2 more answers

Verify that the SI unit of impulse is the same as the SI unit of momentum.

lys-0071 [83]

Maybe this will help you out:

Momentum is calculate by the formula:

$P = mv$

Where:

P = momentum

m = mass

v = velocity

The SI unit:

$mass = kg\\ velocity = \dfrac{m}{s}$

So the unit of momentum would be:

$kg.\dfrac{m}{s}$

Impulse is defined as the change in momentum or how much force changes momentum. It can be calculate with the formula:

I = FΔt

where:

I = impulse

F = Force

Δt = change in time

The SI unit:

F = Newtons (N) or $kg.\dfrac{m}{s^{2} }$

t = Seconds (s)

So the unit of impulse would be derived this way:

I = FΔt

I = $kg.\dfrac{m}{s^{2} }$ x $s$

$\dfrac{kg.m.s}{s^{2}} = \dfrac{kg.m.s}{s.s}$

You can then cancel out one s each from the numerator and denominator and you'll be left with:

$kg.\dfrac{m}{s}$

So then:

Momentum: Impulse

$kg.\dfrac{m}{s}$ $kg.\dfrac{m}{s}$

4 0

4 years ago