The diagram shows two different types of fossils from the

An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 115 m/s squared. What ar

Iteru [2.4K]

The direction of the electric field would be south.

qE/m = 115
 E = 115*m/q 
 = 115 * 9.1 * 10^(-31) / 1.67*10^(-19) 
 = 762.87 * 10^(-12) 
 = 6.27 x 10^-10 N/C

Hope this answers the question. Have a nice day. Feel free to ask more questions.

6 0

3 years ago

Please need help fast

iVinArrow [24]

(a) See graph in attachment

The appropriate graph to draw in this part is a graph of velocity vs time.

In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.

Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at

t = 10 s

v = 15 m/s

(b) $1.5 m/s^2$

The average acceleration of the horse can be calculated as:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval

In this problem,

u = 0

v = 15 m/s

t = 10 s

Substituting,

$a=\frac{15-0}{10}=1.5 m/s^2$

(c) 75 m

For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance travelled

u is the initial velocity

t is the time interval

a is the acceleration

In this problem,

u = 0

t = 10 s

$a=1.5 m/s^2$

Substituting,

$s=0+\frac{1}{2}(1.5)(10)^2=75 m$

(d) See attached graphs

In a uniformly accelerated motion:

- The distance travelled (x) follows the equation mentioned in part c,

$x=ut+\frac{1}{2}at^2$

So, we see that this has the form of a parabola: therefore, the graph x vs t will represents a parabola.

- The acceleration is constant during the motion, and its value is

$a=1.5 m/s^2$ (calculated in part b)

therefore, the graph acceleration vs time will show a flat line at a constant value of 1.5 m/s^2.

6 0

3 years ago

A 2-kg bowling ball is 1 meter off the ground on a post when it falls just before it reaches the ground it is traveling 4.4m/s a

marta [7]

Mechanical energy is conserved

4 0

3 years ago

Read 2 more answers

1. Potential energy and a roller coaster. A 1000-kg roller coaster car moves from point A to B then C.

zalisa [80]

Answer:(a) With our choice for the zero level for potential energy of the car-Earth system when the car is at point B ,

U

B

=0

When the car is at point A, the potential energy of the car-Earth system is given by

U

A

=mgy

where y is the vertical height above zero level. With 135ft=41.1m, this height is found as:

y=(41.1m)sin40.0

0

=26.4m

Thus,

U

A

=(1000kg)(9.80m/s

2

)(26.4m)=2.59∗10

5

J

The change in potential energy of the car-Earth system as the car moves from A to B is

U

B

−U

A

=0−2.59∗10

5

J=−2.59∗10

5

J

(b) With our choice of the zero configuration for the potential energy of the car-Earth system when the car is at point A, we have U

A

=0. The potential energy of the system when the car is at point B is given by U

B

=mgy, where y is the vertical distance of point B below point A. In part (a), we found the magnitude of this distance to be 26.5m. Because this distance is now below the zero reference level, it is a negative number.

Thus,

8 0

2 years ago

Two objects attract each other gravitationally with a force of 2.5 x 10^-10N when they are 0.25 m apart. Their total mass is 4.0

In-s [12.5K]

Answer:

M = 3.9406 kg and m = 0.0594 kg

Explanation:

The gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance that separates them. Mathematically it is expressed as follows:

Fg = (G×M×m)/r² Formula (1)

Where:

Fg is the gravitational force (N)

G is the universal gravitation constant, G = 6.67 × 10⁻¹¹ (N×m²)/kg²

M and m are the masses of the bodies that interact (kg).

r is the distance that separates them (m).

Known Data

Fg = 2.5 × 10⁻¹⁰ N

r = 0.25 m

G = 6.67 × 10⁻¹¹ (N×m²)/kg²

Problem development

We propose 2 equations

M + m = 4kg

M = 4 - m equation (1)

We replace in formula (1)

2.5 × 10⁻¹⁰ = (6.67 × 10⁻¹¹ × M × m)/(0.25)²

2.5 × 10⁻¹⁰ × (0.25)² = (6.67 × 10⁻¹¹ × M × m)

(2.5 × 10⁻¹⁰ × (0.25)²)/(6.67 × 10⁻¹¹) = M × m

M × m = 0.234 equation (2)

We replace M = 4 - m in equation (2)

(4 - m) × m = 0.234

4m - m² = 0.234

m² - 4m + 0.234 = 0 (quadratic equation)

We apply the formula for the quadratic equation and obtain 2 values for m that meet the conditions:

m = 3.9406 kg or m = 0.0594 kg

We replace m in equation (1)

M = 4 - 3.9406 = 0.0594 kg or M = 4 - 0.0594 = 3.9406

To meet the condition that M + m must give 4 kg, one mass must be equal 3.9406 and the other must equal 0.0594, then:

M = 3.9406 kg and m = 0.0594 kg

6 0

3 years ago