HELP ME ASAP PLZ!

car travels 80 meters due north in 12 seconds then the car turns around and travels 30 Mi do South in 4 seconds calculate the av

Zinaida [17]

Answer:

1) 3.1 m/s

2) 7 m/s

Explanation:

Distance due north = 80 m

Distance due south = 30 m

Distance between north and south = (80 - 30) m = 50 m

Total time = (12 + 4) sec = 16 sec

1) Average speed = 50/16 = 3.1 m/s

2) Average velocity = Total distance/total time = (80 + 30) m/16 s = 110/16 = 7 m/s

8 0

4 years ago

A man pushes a lawn mower on a level lawn with a force of 195 N. If 37% of this force is directed downward, how much work is don

densk [106]

Answer:

$W = 1032.6 J$

Explanation:

Net force by which man push the lawn mower is given as

$F = 195 N$

now it is given that 37% of this force is vertically downwards

so we will have

$F_y = 0.37 \times 195$

$F_y = 72.15 N$

now we also know that

$F_x^2 + F_y^2 = F^2$

here we have

$F_x^2 + 72.15^2 = 195^2$

$F_x = 181.2 N$

Now work done by this force to move the lawn mower is given as

$W = F_x . d$

$W = (181.2)(5.7 m)$

$W = 1032.6 J$

7 0

3 years ago

Distance v. Time

Allushta [10]

Answer:

4

(m)

2 ( s )

Explanation:

ok...........

3 0

4 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

What object currently has the most gravitational potential energy?<br><br>A, B, C, or D

eduard

Answer:

A

Explanation:

this because

gravitational potential energy = mass x height x gravitational field strength

so let's assume mass is 2 kg and gravitational field strength is 10 N /kg

so when height is very low, take it as 3 m

gravitational potential energy= 2 x 3 x 10 = 60 j

but when height is 6m

gravitational potential energy = 2 x 6 x 10 = 120 j

so when the height is the greatest, the gravitational potential energy is the highest

so A is the heighest so it has the highest gravitational potential energy.

hope this helps

please mark it brainliest :D

5 0

3 years ago