Ifa cup of coffee is at 90°C and a person with a body temperature of 36°C touches it, how will heat flow between

A 150 g egg is dropped from 3.0 meters. The egg is

Lynna [10]

Answer:

9.2 N, with significant figure rounding (2 s.f.)

Explanation:

This problem can be solved using momentum. The following equation relates momentum (mass & velocity) with force and time:

Note that where v is the final velocity and v₀ is the initial velocity. Δv just means change in velocity.

Mass of the egg is 150 g, but we need to convert to kilograms if we want to use Newtons as a unit. 150 g is equal to 0.15 kg. since 1000 g = 1kg.

m = 0.15 kg

The dropped from 3.0 meters is irrelevant as the question tells us the initial velocity of the egg: 4.4 m/s before it hits the ground.

v₀ = 4.4 m/s [down]

When it comes to a stop, the egg will have a velocity of 0.

v = 0 m/s

The time it takes for the egg to stop is 0.072 seconds.

Δt = 0.072 s

Therefore, if down is positive, then

We round to two significant figures since every quantity has two sig. figs.

We only care about the magnitude, not direction. The answer is 9.2 N.

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5.0

5 0

2 years ago

What is the average horizontal component of force exerted on his feet by the ground during acceleration?

Arte-miy333 [17]

Well you have to think of it like electricity go through your answer closes to that and figure it out

7 0

3 years ago

A 70 Newton dog ran across the yard. What do we know about the dog?

Amiraneli [1.4K]

The dogs acceleration

5 0

3 years ago

1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal.

harkovskaia [24]

Answer:

(a) The net force is 80.394 N

The acceleration of the crate is 0.804 m/s²

(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, $F_y = Fsin \theta = 250sin45 = 176.78 \ N$

Applied force in x-direction, $F_x = Fcos \theta = 250cos45 = 176.78 \ N$

The normal force is calculated as;

N + Fy -W = 0

N = W - Fy

N = (100 x 9.8) - 176.78

N = 980 - 176.78 = 803.22 N

The frictional force is given by;

Fk = μN

Fk = 0.12 x 803.22

Fk = 96.386 N

(a) The net force is given by;

$F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N$

Apply Newton's second law of motion;

F = ma

$a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2$

(b) the velocity of the crate after 5.0 s

$F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s$

7 0

3 years ago

how long does it take adrian to reach his destination if his displacement is 253 meters and he walks north with a velocity of 1.

klemol [59]

Velocity = displacement/time

1.6 = 253/t

t = 158.125

It takes them about 158 seconds

4 0

3 years ago