What forces are acting when you try to crack<br> an egg in your palm(will give brainliest)

A golf club hits a stationary 0.05kg golf ball with and average force of 5.0 x 10^3 newtons accelerating the ball at 44 meters p

maxonik [38]

Answer: The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

Explanation:

Force applied on the golf ball = $5.0\times 10^3 N$

Mass of the ball = 0.05 kg

Velocity with which ball is accelerating = 44 m/s

Time period over which forece applied = t

$f=ma=\frac{m\times v}{t}$

$t=\frac{0.05 kg\times 44m/s}{5.0\times 10^3 N}=4.4\times 10^{-4} seconds$

$Impulse=(force)\times (time)=f\times t = 5.0\times 10^3\times 4.4\times 10^{-4} seconds=2.2$ Newton seconds

The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

7 0

3 years ago

The number of mushrooms needed by a pizza restaurant is given by the

kupik [55]

Answer: 24

Explanation:

Given the following equation:

$g=3n$

Where $g$ is the number of mushrooms in a pizza and $n$ the number of pizzas.

If we know the restaurant will make 8 pizzas ( $n=8$ ), then:

$g=3(8)$

$g=24$ This is the needed number of mushrooms for 8 pizzas

8 0

3 years ago

Lance Armstrong bikes at a constant speed up the Col d’Izoard, a famous mountain pass. Assume his teammates do such a good job r

svetoff [14.1K]

Work done against gravity to climb upwards is always stored in the form of gravitational potential energy

so we can say

$W = mgh$

here h = vertical height raised

so here we know that

$h = 14.1 sin7.3 km$

here we have

$h = 1.79 km$

now from above equation

$W = (83 kg)(9.81 m/s^2)(1.79 \times 10^3 m)$

$W = 1.46 \times 10^6 J$

so work done will be given by above value

7 0

3 years ago

If the mass of an object is 44 kilograms and its velocity is 10 meters per second east, how much Kinetic Energy does it have?

olya-2409 [2.1K]

Given: Mass m = 44 Kg; Velocity v = 10 m/s

Required: Kinetic energy K.E = ?

Formula: K.E = 1/2 mv²

K.E 1/2 (44 Kg)(10 m/s)²

K.E = 2,200 Kg.m²/s²

K.E = 2,200 J Answer is A

8 0

3 years ago

A 2.31 kg rope is stretched between supports 10.4 m apart. If one end of the rope is tweaked, how long will it take for the resu

zlopas [31]

Answer:

t = 0.657 s

Explanation:

First, let's use the appropiate equations to solve this:

V = √T/u

This expression gives us a relation between speed of a disturbance and the properties of the material, in this case, the rope.

Where:

V: Speed of the disturbance

T: Tension of the rope

u: linear density of the rope.

The density of the rope can be calculated using the following expression:

u = M/L

Where:

M: mass of the rope

L: Length of the rope.

We already have the mass and length, which is the distance of the rope with the supports. Replacing the data we have:

u = 2.31 / 10.4 = 0.222 kg/m

Now, replacing in the first equation:

V = √55.7/0.222 = √250.9

V = 15.84 m/s

Finally the time can be calculated with the following expression:

V = L/t ----> t = L/V

Replacing:

t = 10.4 / 15.84

t = 0.657 s

4 0

3 years ago

What forces are acting when you try to crack an egg in your palm(will give brainliest)