One who voluntarily donates tissues/ organs striped

Ivan drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 7 hours. When Ivan drove h

lesya692 [45]

Answer:252 miles

Explanation:

Given

During his way to mountain it took 7 hr to drive

and during his return trip it took 4 hr to return

Let x be the distance between home and mountain

average speed for return is 27 miles per hour faster than his former trip

let v be the speed on his way to mountain thus v+27 is his return speed

thus $7=\frac{x}{v}$ ----1

for return trip

$4=\frac{x}{v+27}$ -----2

divide 1 & 2

$\frac{7}{4}=\frac{x\cdot (v+27)}{v\cdot x}$

$7v=4v+4\cdot 27$

$3v=4\cdot 27$

$v=36 mph$

thus $x=7\times 36=252\ miles$

7 0

3 years ago

a child's toy consists of a piece of plastic attached to a spring. the spring is compressed against the floor a distance of 2.0

Nat2105 [25]

Answer:

1.7N

Explanation:

Force = kx

Where x = spring compression and

K = spring constant

K =85N/m

x = 2.0cm / 100

= 0.02m

Force = 85 x 0.02

= 1.7N

5 0

3 years ago

a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long

Phantasy [73]

Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

$a=\frac{Vf-Vi}{t}$

$-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}$

In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:

$-9.8 =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s$

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

$Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m$

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

$Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2$