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3 years ago

2. How does the mass of an object affect its gravitational pull?

1 answer:

8 0

Objects with more mass has more gravity so, the closer objects are to each other the stronger gravitational pull. Hope this helps

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Physics Homework MathPhys homie if you see this pls help

cluponka [151]

Answer:

1. -8.20 m/s²

2. 73.4 m

3. 19.4 m

Explanation:

1. Apply Newton's second law to the car in the y direction.

∑F = ma

N − mg = 0

N = mg

Apply Newton's second law to the car in the x direction.

∑F = ma

-F = ma

-Nμ = ma

-mgμ = ma

a = -gμ

Given μ = 0.837:

a = -(9.8 m/s²) (0.837)

a = -8.20 m/s²

2. Given:

v₀ = 34.7 m/s

v = 0 m/s

a = -8.20 m/s²

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (34.7 m/s)² + 2 (-8.20 m/s²) Δx

Δx = 73.4 m

3. Since your braking distance is the same as the car in front of you, the minimum safe following distance is the distance you travel during your reaction time.

d = v₀t

d = (34.7 m/s) (0.56 s)

d = 19.4 m

6 0

3 years ago

Your friend, who is in a field 100 meters away from you, kicks a ball towards you with an initial velocity of 16 m/s. Assuming t

LekaFEV [45]

Answer:

Time, t = 5.355 seconds

Explanation:

Given the following data;

Distance = 100 m

Initial velocity = 16 m/s

Deceleration = 1 m/s²

To find the time, we would use the second equation of motion;

But since the ball is decelerating, it's acceleration would be negative.

S = ut + ½at²

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

100 = 16t - 0.5t²

200 = 32t - t²

t² + 32t - 200 = 0

Solving the quadratic equation using the quadratic formula;

The quadratic equation formula is;

$x = \frac {-b \; \pm \sqrt {b^{2} - 4ac}}{2a}$

Substituting into the equation, we have;

$x = \frac {-32 \; \pm \sqrt {32^{2} - 4*1*(-200)}}{2*1}$

$x = \frac {-32\pm \sqrt {1024 - (-800)}}{2}$

$x = \frac {-32 \pm \sqrt {1024 + 800}}{2}$

$x = \frac {-32 \pm \sqrt {1824}}{2}$

$x = \frac {-32 \pm 42.71}{2}$

$x_{1} = \frac {-32 + 42.71}{2}$

$x_{1} = \frac {10.71}{2}$

x1 = 5.355

We do not need the negative value of x, so we proceed.

Therefore, time = 5.355 seconds

3 0

3 years ago

Diane swim 5 kilometers against the current in the same amout of time it took her to swim 15 kilometers with the current . The r

lidiya [134]

Answer:

4 km/hr

Explanation:

suppose 's' is Diane's speed with no current.

't' represents time in hrs.

Using the formula:

Distance = speed 's' x time 't'

-> when she swims against the current, equation will be,

5= (s-2)t

t= 5/(s-2)

->when she was swimming with the current, equation is,

15= (s+2) t

t= 15/(s+2)

equating eq(1) and (2)

5/(s-2) = 15/(s+2)

5s + 10 = 15s - 30

40= 10s

s= 40/10

s=4

Therefore, if there were no current, her speed is 4km/hr

5 0

3 years ago

A specialist who treats skin conditions and diseases is a

s2008m [1.1K]

A doctor who specializes in the area of skin health is a dermatologist

5 0

3 years ago

What is a volcanically active area that is not near a tectonic plate boundary?

belka [17]

It a blue tectonic movement

8 0

3 years ago