I need 4 answers for Earth's layers and 4 answers for the Descriptions for Earth's layers graphic organizer.

A car of mass m push = 1200 kg is capable of a maximum acceleration of 6.00 m / s 2 . If this car is required to push a stalled

motikmotik

1)

first you find the maxium force that the car can produce.

f=ma

Fmax=(1100kg)(6m/s^2)

then use f = ma again to find the accel with the passengers

Fmax=(1100kg +1650kg)(a)

=> a = (1100kg)(6m/s^2)/( 1100kg +1650kg)

= 2.4 m/s^2

4 0

2 years ago

1. Cam Newton can sprint 40 meters in 5.79 seconds! How fast can he run?

Vesnalui [34]

This is a Physics question where we need to figure out how many meters Cam can run per second. To figure this out we divide the distance by the change in time.

40/5.79 = 6.9 meters per second approximately.

3 0

2 years ago

Read 2 more answers

A spotlight on the ground shines on a wall 12 meters away. If a man 2 meters high walks away from the spotlight towards the wall

nordsb [41]

Answer: The length of the shadow on the wall is decreasing by 0.6m/s

Explanation:

the specified moment in the problem, the man is standing at point D with his head at point E.

At that moment, his shadow on the wall is y=BC.

The two right triangles ΔABC and ΔADE are similar triangles. As such, their corresponding sides have equal ratios:

ADAB=DEBC

8/12=2/y,∴y=3 meters

If we consider the distance of the man from the building as x then the distance from the spotlight to the man is 12−x.

(12−x) /12=2/y

1− (1 /12x )=2 × 1/y

Let's take derivatives of both sides:

−1 / 12dx = −2 × 1 / y^2 dy

Let's divide both sides by dt:

−1/12⋅dx/dt=−2/y^2⋅dy/dt

At the specified moment:

dxdt=1.6 m/s

y=3

Let's plug them in:

−1/121.6) = - 2/9 × dy/dt

dy/dt = 1.6/12 ÷ 2/9

dy/dt = 1.6/12 × 9/2

dy/dt = 14.4/24 = 0.6m/s

5 0

3 years ago

ΔP = 1.88 x 10^4 Pa. Use this answer to estimate the volume flow rate of blood from the head to the feet of a six-foot-tall pers

Sveta_85 [38]

Answer: $3765.66 \frac{m^{3}}{s}$

Explanation:

We can solve this problem using the <u>Poiseuille equation</u>:

$Q=\frac{\pi r^{4}\Delta P}{8\eta L}$

Where:

$Q$ is the Volume flow rate

$r=23 cm \frac{1 m}{100 cm}=0.23 m$ is the effective radius

$L=6 ft \frac{0.3048 m}{1 ft}=1.8288 m$ is the length

$\Delta P=1.88(10)^{4} Pa$ is the difference in pressure

$\eta=3(10)^{-3} Pa.s$ is the viscosity of blood

Solving:

$Q=\frac{\pi (0.23 m)^{4}(1.88(10)^{4} Pa)}{8(3(10)^{-3} Pa.s)(1.8288 m)}$

$Q=3765.66 \frac{m^{3}}{s}$

7 0

2 years ago

Light from a distant star is collected by a concave mirror that has a radius of curvature of 150 cm. How far from the mirror is

andrey2020 [161]

The focal point of a mirror is half of the radius of curvature. We can use the formula, 1/f = 1/v + 1/u where f is the focal length , v is the image distance and u is object distance The distance of the star is assumed to be incredibly far away and 1 divided by a really big number is approximately zero, thus; 1/f = 1/v = 1/75 = 1/v therefore; the image is formed 75 cm infront of the mirror

4 0

3 years ago