Ask question

Ask question

All categories

3 years ago

13

A gymnast jumps directly up onto a beam that is 1.5 m high. From the beam, she jumps 0.5 m straight up and lands back on the bea

m in the same spot. What is the total distance she traveled?
A. 0.5 m
B. 1.0 m
C. 1.5 m
D. 2.5 m

1 answer:

tatiyna3 years ago

5 0

Answer:

D. 2.5 m

Explanation:

1.5 m up to the beam + 0.5 m up + 0.5 m down = 2.5 m total

You might be interested in

Question 3 of 10

Eva8 [605]

Answer:

D. Newton's second law

Explanation:

Newton's second law of motion states that force of an object is a product of its mass and its acceleration.

Mathematically, F= ma where m is mass and a is acceleration

So from the statement above : The acceleration of an object is proportional to the force applied to it and inversely proportional to its mass , it can be seen from the formula variation as;

F= ma -----making a the subject of the formula

a= F/ m

a= 1/m * F --------- a is inversely related to m as you can see from 1/m but directly related to F hence;

Increase in mass with the same force applied causes the body to accelerate slower where as when force increases, the body accelerates faster.

5 0

2 years ago

If 50 km thick crust having an average density of 3.0 g/cm3 has a surface elevation of 2.5 km above sea level, what would you pr

RUDIKE [14]

Answer:

To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.

The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.

Thus let the density of the material be Pm

50*3= 47.5*Pm

Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube

Thus with an average density of 2.8gram per centimeter cube

50*2.8= (50-x)*3.16

(50-x)= (50*2.8)/3.16

50-x=44.3

x=50-44.3= 5.7

Explanation:

To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.

The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.

Thus let the density of the material be Pm

50*3= 47.5*Pm

Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube

Thus with an average density of 2.8gram per centimeter cube

50*2.8= (50-x)*3.16

(50-x)= (50*2.8)/3.16

50-x=44.3

x=50-44.3= 5.7

5 0

3 years ago

A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 836 N.

kotegsom [21]

<span>A= a=99/85.3
B= a=-54/85.3
C=The acceleration is smaller.
</span>

7 0

3 years ago

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the <u>initial point</u> is 19820 J

2) The potential energy at <u>point A</u> is 19620 J

3) The kinetic energy at <u>point B</u> is 10010 J

4) The potential energy at <u>point C</u> is zero

5) The kinetic energy at <u>point C</u> is 19820 J

6) The velocity of the roller coaster at <u>point C</u> is 19.91 m/s

1) The total energy for the roller coaster at the <u>initial point</u> can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The<em> total energy</em> is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the <u>initial point</u> is 19820 J.

2) The <em>potential energy</em> at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at <u>point A</u> is 19620 J.

3) The <em>kinetic energy</em> at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at <u>point B</u> is 10010 J.

4) The <em>potential energy</em> at <u>point C</u> is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The <em>kinetic energy</em> of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at <u>point C</u> is 19820 J.

6) The <em>velocity</em> of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at <u>point C</u> is 19.91 m/s.

Read more here:

brainly.com/question/21288807?referrer=searchResults

I hope it helps you!

3 0

3 years ago

Find the net charge of a system consisting of 6.25×10^6 electrons and 7.75×10^6 protons. Express your answer using three signifi

jok3333 [9.3K]

Answer:

2.40 x 10⁻¹³ C

Explanation:

$n_{e}$ = number of electrons = 6.25 x 10⁶

$q_{e}$ = charge on electron = - 1.6 x 10⁻¹⁹ C

$n_{p}$ = number of protons = 7.75 x 10⁶

$q_{p}$ = charge on proton = 1.6 x 10⁻¹⁹ C

Net charge is given as

Q = $n_{e}$ $q_{e}$ + $n_{p}$ $q_{p}$

Q = (- 1.6 x 10⁻¹⁹) (6.25 x 10⁶) + (1.6 x 10⁻¹⁹) (7.75 x 10⁶)

Q = 2.40 x 10⁻¹³ C

6 0

3 years ago