During the _____ collision, the vehfirst second third fourthicle comes to an abrupt stop.

A car going initially with a velocity 13.5 m/s accelerates at a rate of 1.9 m/s for 6.2 s. It then accelerates at a rate of-1.2

yanalaym [24]

Answer:

a) Maximum speed = 25.28 m/s

b) Total time = 27.27 s

c) Total distance traveled = 402.43 m

Explanation:

a) Maximum speed is obtained after the end of acceleration

v = u + at

v = 13.5 + 1.9 x 6.2 = 25.28 m/s

Maximum speed = 25.28 m/s

b) We have maximum speed = 25.28 m/s, then it decelerates 1.2 m/s² until it stops.

v = u + at

0 = 25.28 - 1.2 t

t = 21.07 s

Total time = 6.2 + 21.07 = 27.27 s

c) Distance traveled for the first 6.2 s

s = ut + 0.5 at²

s = 13.5 x 6.2 + 0.5 x 1.9 x 6.2² = 120.22 m

Distance traveled for the second 21.07 s

s = ut + 0.5 at²

s = 25.28 x 21.07 - 0.5 x 1.2 x 21.07² = 282.21 m

Total distance traveled = 120.22 + 282.21 = 402.43 m

5 0

3 years ago

Read 2 more answers

You serve a volleyball with a mass of 1.4 kg. The ball leaves with a speed of 13 m/s. Calculate KE

NemiM [27]

Answer:

118.3 J

Explanation:

Givens:

m = 1.4 kg

V = 13 m/s

Formula for kinetic energy:

KE = (1/2)*(m)*(v)^2

KE = .5*(1.4 kg)*(13 m/s)^2

KE 118.3 J

J = Joules

7 0

3 years ago

A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f

VashaNatasha [74]

Answer:

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 = X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + $(\frac{L-y}{2})$ ) g sinФ - F2 x ( $\frac{L}{2}$ ) x gsinФ = 0

plug in the equations of F1 and F2 into the above equation and after simplification, we get:

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . 5/9 x w

Hence,

$(L^{2} - y^{2} )$ = $\frac{5L^{2} }{9}$

$\frac{L^{2} - y^{2} }{L^{2} }$ = $\frac{5}{9}$

Taking $L^{2}$ common and solving for $\frac{y}{L}$ , we will get

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

8 0

3 years ago

When plugging in metric facts, always remember that 1 big unit = # small units. Fill in these facts: 1.________ s= ________μs 2.

balandron [24]

Answer:

1. 1 s = 1 x 10⁶ μs

2. 1 g = 0.001 kg

3. 1 km = 1000 m

4. 1 mm = 1 x 10⁻³ m

5. 1 mL = 1 x 10⁻³ L

6. 1 g = 100 dg

7. 1 cm = 1 x 10⁻² m

8. 1 ms = 1 x 10⁻³ s

Explanation:

1.

1 x 10⁻⁶ s = 1 μs

(1 x 10⁻⁶ x 10⁶) s = 1 x 10⁶ μs

1 s = 1 x 10⁶ μs

2.

1000 g = 1 kg

1 g = 1/1000 kg

1 g = 0.001 kg

3.

1 km = 1000 m

4.

1 mm = 1 x 10⁻³ m

5.

1 mL = 1 x 10⁻³ L

6.

1 x 10⁻² g = 1 dg

(1 x 10⁻² x 10²) g = 1 x 10² dg

1 g = 100 dg

7.

1 cm = 1 x 10⁻² m

8.

1 ms = 1 x 10⁻³ s

4 0

3 years ago

A 200-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevatio

egoroff_w [7]

Answer:

665 ft

Explanation:

Let d be the distance from the person to the monument. Note that d is perpendicular to the monument and would make 2 triangles with the monuments, 1 up and 1 down.

The side length of the up right-triangle knowing the other side is d and the angle of elevation is 13 degrees is

$dtan13^0 = 0.231d$

Similarly, the side length of the down right-triangle knowing the other side is d and the angle of depression is 4 degrees

$dtan4^0 = 0.07d$

Since the 2 sides length above make up the 200 foot monument, their total length is

0.231d + 0.07d = 200

0.301 d = 200

d = 200 / 0.301 = 665 ft

7 0

3 years ago