A car veers off course and runs straight into a brick wall. This is an example

a baby carriage is sitting at the top of a hill that is 21m high. The carriage with the baby weighs 12kg. The carriage has _____

Ket [755]

Height=h=21m
Mass=m=12kg

$\\ \rm\longmapsto P.E=mgh$

$\\ \rm\longmapsto P.E=12(10)(21)$

$\\ \rm\longmapsto P.E=120(21)$

$\\ \rm\longmapsto P.E=2520J$

8 0

3 years ago

4. Describe how the velocity of an object changes if it undergoes uniformly acceleration motion. Can its direction change?

valentinak56 [21]

Answer:

n the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.

In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.

Explanation:

Velocity is a vector therefore it has magnitude and direction, a change in either of the two is the consequence of an acceleration on the system.

In the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.

$a_{t}$ = (v₂-v₁)/Δt

In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.

$a_{c}$ = v2/R

In the general case, both the module and the address change

a = Ra ( a_{t}^2 + a_{c}^2)

4 0

3 years ago

If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun

zvonat [6]

Answer:

H = 1/2 g t^2 where t is time to fall a height H

H = 1/8 g T^2 where T is total time in air (2 t = T)

R = V T cos θ horizontal range

3/4 g T^2 = V T cos θ 6 H = R given in problem

cos θ = 3 g T / (4 V) (I)

Now t = V sin θ / g time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V) from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97 6 within rounding

3 0

3 years ago

A man works on a striaght road from his home to market 2.5km away dwith a speed of 5km/h.Finding the market closed, he instantly

umka21 [38]

Answer:

5.625km/h

Explanation:

We are given that

Distance between home and market, d=2.5 km

Speed, v1=5km/h

Speed, v2=7.5 km/h

We have to find the average speed of the man over the interval of time 0 to 40min.

Time, $t=\frac{distance}{speed}$

Using the formula

$t_1=\frac{2.5}{5}=0.5 h=0.5\times 60=30 min$

1 hour =60 min

$t_2=\frac{2.5}{7.5}=\frac{1}{3}hour=\frac{60}{3}=20min$

Distance traveled by man in 10 min with speed 7.5 km/h= $\frac{2.5}{20}\times 10=2.5/2km$

Therefore,

Total distance covered=2.5+2.5/2=3.75 km

Time=40 min=40/60=2/3 hour

Average speed= $\frac{total\;distance}{total\;time}$

Average speed= $\frac{3.75}{2/3}=5.625km/h$

7 0

3 years ago

H H H O H O N C C N C C H O H H O H H C H H H ALANINE GLYCINE

Trava [24]

i’m not 100 percent sure but I need points to ask questions good luck th

6 0

2 years ago