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2 years ago

11

A quarterback throws a football down the field to the receiver. What type(s) of energy does the football possess? Check all that

apply. chemical electrical kinetic mechanical potential

2 answers:

KengaRu [80]2 years ago

7 0

Answer: kinetic, mechanical, potential

Explanation: Just answered the question

zloy xaker [14]2 years ago

6 0

Answer: kinetic , potential, mechanical

Explanation:

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Which of the answer choices is a true statement? Select all that apply. Metallic elements usually exist as molecules, combining

Pani-rosa [81]

Metallic elements can exist on their own as individual atoms.

Nonmetals usually exist as molecules, combining with atoms of themselves.

Nonmetals can exist on their own as individual atoms.

Explanation:

Metallic elements can exist on their own as individual atoms without combining with another atom. They are stable that way. For example gold, copper, and silver.

Non-metals are usually found as molecules in combined states. The molecules are bounded to each other through covalent bonds. Examples are oxygen gas, nitrogen gas e.t.c. When the two atoms combine, they share their electrons to for homonuclear molecules.

Non-metals like the noble gases exists on their own as individual atoms. They are mono-atomic gases and are stable in nature.

learn more:

Metals and non - metals brainly.com/question/2758034

#learnwithBrainly

6 0

2 years ago

A woman is applying 300N/m2 of pressure on to door with her hand. Her hand has area of 0.02m2. Work out the force being applied

never [62]

Answer:

6N

Explanation:

Given parameters:

Pressure applied by the woman = 300N/m²

Area = 0.02m²

Unknown:

Force applied = ?

Solution:

Pressure is the force per unit area on a body

Pressure = $\frac{force}{area}$

Force = Pressure x area

Force = 300 x 0.02 = 6N

8 0

2 years ago

In which landscape region of new york is the most resistant bedrock found?

Nutka1998 [239]

<span>The Adirondack Mountains, Taconic Mountains, and the Hudson Highlands have the most resistant bedrock.</span>

7 0

2 years ago

Read 2 more answers

 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer

Andrew [12]

Answer:

a) the mug hits the floor 0.7425m away from the end of the bar. b) |V|=5.08m/s θ= -72.82°

Explanation:

In order to solve this problem, we must first start by doing a drawing of the situation. (see attached picture).

a)

From the drawing we can see that we are dealing with a two dimensions movement problem. So in order to find out how far away from the bar the mug will fall, we need to start by finding how long it will take the mug to be in the air, so we analyze the vertical movement of the mug.

In order to find the time we need to use the following formula, which contains the data we know:

$y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}$

we know that $y_{f}=0$ and that $v_{y0}=0$ as well, so the formula is simplified to:

$0=y_{0}+\frac{1}{2}at^{2}$

we can now solve this for t, so we get:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

we know that $y_{0}=1.20m$ and that $a=g=-9.8m/s^{2}$

the acceleration of gravity is negative because the mug is moving downwards. So we substitute them into the given formula:

$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

which yields:

t=0.495s

we can now use this to find the horizontal distance the mug travels. We know that:

$V_{x}=\frac{x}{t}$

so we can solve this for x, so we get:

$x=V_{x}t$

and we can now substitute the values we know:

x=(1.5m/s)(0.495s)

which yields:

x=0.7425m

b) Now that we know the time it takes the mug to hit the floor, we can use it to find the final velocity in the y-direction by using the following formula:

$a=\frac{v_{f}-v_{0}}{t}$

we know the initial velocity in the vertical direction is zero, so we can simplify the formula:

$a=\frac{v_{f}}{t}$

so we can solve this for the final velocity:

$V_{yf}=at$

in this case the acceleration is the same as the acceleration of gravity (which is negative) so we can substitute that and the time we found on the previous part to get:

$V_{yf}=(-9.8m/s^{2})(0.495s)$

which yields:

$V_{yf}=-4.851m/s$

so now we know the components of the final velocity, which are:

$V_{xf}=1.5m/s$ and $V_{yf]=-4.851m/s$

so now we can find the speed by determining the magnitude of the vector, like this:

$|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}$

so we get:

$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

which yields:

|V|=5.08m/s

now, to find the direction of the impact, we can use the following equation:

$\theta = tan^{-1} (\frac{V_{y}}{V_{x}})$

so we get:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

which yields:

$\theta = -72.82^{o}$

4 0

2 years ago

A 2-kg cart, traveling on a horizontal air track with a speed of 3 m/s, collides with a stationary 4-kg cart. The carts stick to

daser333 [38]

Answer:

Magnitude of impulse, |J| = 4 kg-m/s

Explanation:

It is given that,

Mass of cart 1, $m_1=2\ kg$

Mass of cart 2, $m_2=4\ kg$

Initial speed of cart 1, $u_1=3\ m/s$

Initial speed of cart 2, $u_2=0$ (stationary)

The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.

$m_1u_1+m_2u_2=(m_1+m_2)V$

$V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$V=\dfrac{2\times 3}{(2+4)}$

V = 1 m/s

The magnitude of the impulse exerted by one cart on the other is given by:

$J=F\times t=m(V-u)$

$J=m(V-u)$

$J=2\times (1-3)$

J = -4 kg-m/s

or

|J| = 4 kg-m/s

So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.

8 0

2 years ago