Ask question

Ask question

All categories

2 years ago

7

3) A defense football player on one team tackles the other team’s quarterback, who is running down the field. The quarterback is

pushed to the ground by the defensive player. 4) apply Newton’s first law to explain what happened?

1 answer:

alisha [4.7K]2 years ago

5 0

Newton's first law says that an object at rest tends to stay at rest while an object in motion stays in motion at a constant velocity unless acted upon by an outside force so the amount of force behind the defensive football player (N) was greater than the quarterback's so he was able to over power him which is also called unbalanced forces

You might be interested in

A submerged submarine alters its buoyancy so that it initially accelerates upward at 0.325 m/s^2. What is the submarine's averag

scoundrel [369]

<span>You are given a submerged submarine accelerating upward at 0.325 m/s</span>² and the density of sea water is 1.025x10³ kg/m³. The submarine's average density at this time is 22 kg/m³.

7 0

2 years ago

Two thermometers are calibrated, one in degrees Celsius and the other in degrees Fahrenheit.

sdas [7]

Answer:

The temperature is 233.15 K

Explanation:

Recall the formula to convert degree Celsius (C) into Fahrenheit (F):

$\frac{9}{5} C+32=F$

So if we want the value of degree C to be the same as the value of the degree F, we want the following: C = F

which replacing F with C on the right hand side of the equation above, allows us to solve for C:

$\frac{9}{5} C+32=F\\\frac{9}{5} C+32=C\\\frac{9}{5} C-\frac{5}{5} C =-32\\\frac{4}{5} C==32\\C= \frac{-32\,*\,5}{4} \\C=-40$

This means that -40°C = -40°F

And this temperature in Kelvin is:

-40°C + 273.15 = 233.15 K

8 0

2 years ago

A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th

ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = $9.8\ m/s^2$

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

Kinetic energy K.E = $\frac{1}{2}mv^2$ .

From above we know that -

Kinetic energy at the bottom of the cliff = potential energy at a height h

$\frac{1}{2}mv^2=\ mgh$

⇒ $v^2=\ 2gh$

⇒ $h=\ \frac{v^2}{2g}$

⇒ $h=\ \frac{(24.2)^2}{2\times 9.8}$

⇒ $h=\ 29.88\ m$

Hence, the height of the cliff is 29.88 m

5 0

2 years ago

In which of the following examples does the object have both kinetic and potential energy? Select all that apply.

notsponge [240]

I believe the answer is H for when you bounce it, it has stress when it hits the floor and then goes up giving it kinetic

6 0

2 years ago

Read 2 more answers

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

5 0

2 years ago