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3 years ago

Please I really need help on this

Physics

2 answers:

mylen [45]3 years ago

6 0

It would be that all of the elements in the column have similar chemical properties. I hope that this helped you out hon :)

taurus [48]3 years ago

3 0

The secondgfgghhhhhh

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A heater is being designed that uses a coil of 14-gauge nichrome wire to generate 300 W using a voltage of V = 110 V . How long

Mazyrski [523]

Answer:

The length of the wire is 83.2 m.

Explanation:

Given that,

Power of heater, P = 300 W

Voltage, V = 110 V

The resistivity of nichrome wire is, $\rho=100\times 10^{-8}\ \Omega-m$

The electric power of a wire is given by :

$P=\dfrac{V^2}{R}\\\\R=\dfrac{V^2}{P}\\\\R=\dfrac{(110)^2}{300}\\\\R=40.34\ \Omega$

Area of cross section of the wire is :

$A=\pi r^2\\\\A=\pi (0.000815)^2\\\\A=2.08\times 10^{-6}\ m^2$

Resistance of a material is given by :

$R=\rho \dfrac{L}{A}\\\\L=\dfrac{RA}{\rho}\\\\L=\dfrac{40\times 2.08\times 10^{-6}}{10^{-6}}\\\\L=83.2\ m$

So, the length of the wire is 83.2 m. Hence, this is the required solution.

5 0

3 years ago

Erase all the trajectories, and fire the pumpkin vertically again with an initial speed of 14 m/s. As you found earlier, the max

yanalaym [24]

Answer:

$\theta=39.49^{\circ}$

Explanation:

Maximum height of the pumpkin, $H_{max}=9.99\ m$

Initial speed, v = 22 m/s

We need to find the angle with which the pumpkin is fired. the maximum height of the projectile is given by :

$H_{max}=\dfrac{v^2\ sin^2\theta}{2g}$

On rearranging the above equation, to find the angle as :

$\theta=sin^{-1}(\dfrac{\sqrt{2gH_{max}}}{v})$

$\theta=sin^{-1}(\dfrac{\sqrt{2\times 9.8\times 9.99}}{22})$

$\theta=39.49^{\circ}$

So, the angle with which the pumpkin is fired is 39.49 degrees. Hence, this is the required solution.

8 0

3 years ago

The escape speed from an object is v2 = 2GM/R, where M is the mass of the object, R is the object's starting radius, and G is th

Rom4ik [11]

Answer:

Approximate escape speed = 45.3 km/s

Explanation:

Escape speed

$v=\sqrt{\frac{2GM}{R}}$

Here we have

Gravitational constant = G = 6.67 × 10⁻¹¹ m³ kg⁻¹ s⁻²

R = 1 AU = 1.496 × 10¹¹ m

M = 2.3 × 10³⁰ kg

Substituting

$v=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 2.3\times 10^{30}}{1.496\times 10^{11}}}=4.53\times 10^4m/s=45.3km/s$

Approximate escape speed = 45.3 km/s

6 0

3 years ago

A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 4.15 m and mass 7.98 kg, find

Ludmilka [50]

Answer:

4.535 N.m

Explanation:

To solve this question, we're going to use the formula for moment of inertia

I = mL²/12

Where

I = moment of inertia

m = mass of the ladder, 7.98 kg

L = length of the ladder, 4.15 m

On solving we have

I = 7.98 * (4.15)² / 12

I = (7.98 * 17.2225) / 12

I = 137.44 / 12

I = 11.45 kg·m²

That is the moment of inertia about the center.

Using this moment of inertia, we multiply it by the angular acceleration to get the needed torque. So that

τ = 11.453 kg·m² * 0.395 rad/s²

τ = 4.535 N·m

8 0

3 years ago

A jet, sitting on the runway, takes off and accelerates at 8.0 m/s for 16s. How far did the jet travel down the runway?

nydimaria [60]

Answer:

2.4 m/s". 1

Explanation:

A jet with mass m = 8 x 10* kg jet accelerates down the runway for takeoff at 2.4 m/s". 1

8 0

3 years ago