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2 years ago

13

Determine (a) the starting height, (b) the time to hit the ground, and (c) the velocity when it hits the ground for an object sh

ot horizontally with a speed of 8 m/s that lands 10 m away.

1 answer:

Salsk061 [2.6K]2 years ago

8 0

Answer:

a

Explanation:

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When did the object have the highest average

miv72 [106K]

Answer:

Between 120 and 180 seconds

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3 years ago

Both a 3 kg book and 5 kg bowling ball are sitting still on the floor. What is true about the kinetic energy of the objects? Exp

Charra [1.4K]

Answer:

The kinetic energy for both objects is the same.

Explanation:

While in other cases the kinetic energies of two objects that have different masses might be different depending on their velocities, in this case both the 3 kg book and 5 kg bowling ball have the same kinetic energy.

This is because kinetic energy is calculated using the formula: K = 1/2 * m * v^2, where m represents the mass and v represents the velocity of the object.

Since the book and the bowling ball are sitting still on the floor, their velocities are zero. Hence, when we plug in 0 for velocity into the equation for kinetic energy, we will get that the kinetic energy is 0 for the book and the bowling ball.

Hope this helps!

6 0

3 years ago

What is the Sl unit of Newton ?............

kenny6666 [7]

Answer:

<em>N</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>SI</em><em> </em><em>unit</em><em> </em><em>of</em><em> </em><em>Newton</em>

5 0

3 years ago

Read 2 more answers

Alarge plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 MPa m1/2. If the plate is expos

Korvikt [17]

Answer:

minimum length of a surface crack is 18.3 mm

Explanation:

Given data

plane strain fracture toughness K = 82.4 MPa m1/2

stress σ = 345 MPa

Y = 1

to find out

the minimum length of a surface crack

solution

we will calculate length by this formula

length = 1/π ( K / σ Y)²

put all value

length = 1/π ( K / σ Y)²

length = 1/π ( 82.4 $10^{3/2}$ / 345× 1)²

length = 18.3 mm

minimum length of a surface crack is 18.3 mm

4 0

3 years ago

The orbital radius of an electron in a hydrogen atom is 0.846 nm. What is its de Broglie wavelength?

kotykmax [81]

Answer:

The value is $\lambda = 1.329 *10^{-9} \ m$

Explanation:

From the question we are told that

The orbital radius is $r = 0.846nm = 0.846 *10^{-9} \ m$

Generally the de Broglie wavelength is mathematically represented as

$\lambda = \frac{2 * \pi r}{4}$

substituting values

$\lambda = \frac{ 2 * 3.142 * 0.846 *10^{-9}}{4}$

$\lambda = 1.329 *10^{-9} \ m$

6 0

3 years ago