A girl travels 20 miles on her bike, the trip takes two hours. express her speed in miles/hr

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.

Unlimited, ad-free access to all of the questions with Brainly Plus

START 7 DAY FREE TRIAL

Click to let others know, how helpful is it

5.0

5 0

3 years ago

When performing an.experiment similar to Millikan's oil drop, a student measured the following load magnitudes: 3.26x10 ^-19 C 5

torisob [31]

Answer:

Explanation:

We put the charges in the ascending order as follows

1.53 P

3.26 P

4.66 P

5.09 P

6.39 P

where P is equal to 10⁻¹⁹

we round off given charges as follows

1.53 P → 1.6 P

3.26 P → 3.2 P

4.66 P → 4.8 P

5.09 P → 4.8 P

6.39 P → 6.4 P

We see that 2 nd to 4 th charges are integral multiples of first charge . That means these charges are supposed to be made of combination of first charge . So first charge appears to be minimum possible charge .

Hence this charge may exist on single electron.

8 0

3 years ago

In a nuclear fusion reaction two 2H atoms are combined to produce one 4He.

Mrrafil [7]

Answer:

a)= 0.025602u

b) = 23.848MeV

c) N = 1.546 × 10¹³

Explanation:

The reaction is

²₁H + ²₁H ⇄ ⁴₂H + Q

a) The mass difference is

Δm = 2m(²₁H) - m (⁴₂H)

= 2(2.014102u) - 4.002602u

= 0.025602u

b) Use the Einstein mass energy relation ship

The enegy release is the mass difference times 931.5MeV/U

E = (0.025602) (931.5)

= 23.848MeV

c)

the number of reaction need per seconds is

N = Q/E

= 59W/ 23.848MeV

$= \frac{59}{(23.848 \times 10^6 )(1.6 \times 10^1^9) } \\\\= 1.546 \times 10^1^3$

N = 1.546 × 10¹³

5 0

3 years ago

I need help with this question how to solve it for Brass and Cooper

Ksenya-84 [330]

Take into account that density and relative density are given by:

$\begin{gathered} \text{density}=\text{ mass/volume} \\ \text{relative density = density/density of water} \end{gathered}$

Take into account that the volume associated to each of the given sustances in the table is determined by the Level Difference (because it is the change in the volume of the water of the recipient in which the substance is immersed).

The density of water in kg/m^3 is 1000 kg/m^3.

Due to the density must be given in kg/m^3, it is necessary to express the volumes of the table in m^3 and mass in kg, then, consider the following conversion factor:

1 m^3 = 1000000 ml

1 kg = 1000 g

Then, you obtain the following results:

Brass:

$\begin{gathered} 53.2g\cdot\frac{1kg}{1000g}=0.0532kg \\ 6ml\cdot\frac{1m^3}{1000000ml}=0.000006m^3 \\ \text{density}=\frac{0.0532kg}{0.000006m^3}\approx8866.67\frac{kg}{m^3} \\ \text{relative density=}\frac{(\frac{8866.66kg}{m^3})}{(1000\frac{kg}{m^3})}\approx8.87 \end{gathered}$

Cooper:

$\begin{gathered} 57.4g=0.0574kg \\ 6ml=0.000006m^3 \\ \text{density}=\frac{0.0574kg}{0.000006m^3}\approx9566.67\frac{kg}{m^3} \\ \text{relative density=}\frac{\frac{9566.67kg}{m^3}}{1000kg}=9.57 \end{gathered}$

3 0

1 year ago

A box sits at rest on a rough 33° inclined plane. Draw the free-body diagram, showing all the forces

Cloud [144]

(a) The free body of all the forces include, frictional force, weight of the box acting perpendicular and another acting parallel to the plane.

(b) When the box is sliding down, the frictional force acts towards the right.

(c) When the box slides up, the direction of the frictional force changes, it acts towards the left.

<h3>Free body diagram</h3>

The free body diagram of all the forces on the box is obtained by noting the upward force and downward forces on the box as shown below;

/ W2

Ф → Ff

↓W1

where;

Ff is the frictional force resisting the down motion of the box
W1 is the perpendicular component of the box weight = Wcos(33)
W2 is the parallel component of the box weight = Wsin(33)

(b) When the box is sliding down, the frictional force acts towards the right.

(c) When the box slides up, the direction of the frictional force changes, it acts towards the left.

Learn more about free body diagram of inclined objects here: brainly.com/question/4176810

8 0

2 years ago