Ask question

Ask question

All categories

3 years ago

8

A man throws a football straight into the air. As it rises, it slows down. Which

1 answer:

mario62 [17]3 years ago

8 0

Answer:

if it is slowing down than it is gaining potential energy

Explanation:

You might be interested in

How far will a car travel it is traveling at 60 mph for 2 hours

Molodets [167]

120 miles because of a car is going at 60 miles per 1 hour, 2 hours would result in 120 miles

3 0

3 years ago

Read 2 more answers

A 0.0450-kg golf ball initially at rest is given a speed of 25.0 m/s when a club strikes. If the club and ball are in contact fo

LuckyWell [14K]

Answer:

Average force, F = 562.5 N

Explanation:

Mass of the golf ball, m = 0.045 kg

Initially, it is at rest, u = 0

Final speed of the ball, v = 25 m/s

The club and the ball are in contact for, $t=2\ ms=2\times 10^{-3}\ s$

We need to find the average force acting on the ball. It can be calculated using the formula as :

$F\times t=m(v-u)$

$F=m\dfrac{v-u}{t}$

$F=0.045\times \dfrac{25}{2\times 10^{-3}}$

F = 562.5 N

So, the average force acting on the ball is 562.5 N. Hence, this is the required solution.

5 0

3 years ago

When a 20-V emf is placed across two resistors in series, a current of 2.0 A is present in each of the resistors. When the same

ehidna [41]

Answer:

7.24 ohm

Explanation:

Let R1 and R2 are resistance of two resistors.

Emf=E=20 V

Current,I=2 A

Current,I'=10 A

We have to find the magnitude of the greater of the two resistances.

In series

$R=R_1+R_2$

$V=IR$

By using the formula

$20=2(R_1+R_2)$

$R_1+R_2=\frac{20}{2}=10$ ...(1)

In parallel

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R}=\frac{R_2+R_1}{R_1R_2}$

$R=\frac{R_1R_2}{R_1+R_2}$

$20=10(\frac{R_1R_2}{R_1+R_2}$

$2=\frac{R_1R_2}{10}$

$R_1R_2=20$

$R_2=\frac{20}{R_1}$

Substitute the value

$\frac{20}{R_1}+R_1=10$

$R^2_1+20=10R_1$

$R^2_1-10R_1+20=0$

$R_1=\frac{10\pm\sqrt{(-10)^2-4(20)}}{2}$

By using quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$R_1=\frac{10\pm 2\sqrt 5}{2}$

$R_1=\frac{10+2\sqrt 5}{2}=5+\sqrt 5=7.24 ohm$

$R_1=\frac{10-2\sqrt 5}{2}=2.76 ohm$

Substitute the value

$R_2=\frac{20}{7.24}=2.76 ohm$

$R_2=\frac{20}{2.76}=7.24 ohm$

Hence, the magnitude of the greater of the two resistance=7.24 ohm

8 0

3 years ago

Read 2 more answers

What is the oxidation state of beryllium

Paha777 [63]

<span> Beryllium has an exclusive </span>+2<span> oxidation state in all of its compounds</span>

3 0

3 years ago

What element has 3 valence electrons and 4 energy levels

Tems11 [23]

Gallium is the answer

6 0

3 years ago