Ground water erosion can result in the formation of stalactite and stalagmite in caves.true or false

A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a

Svet_ta [14]

A) The car overtakes the truck after 7.56 s

B) Initial distance between car and truck: 37.1 m

C) Speed of the truck: 15.9 m/s, speed of the car: 25.7 m/s

D) See graph in attachment

Explanation:

A)

The truck starts from rest and has a constant acceleration, so its position at time t can be written as

$x_t(t)=d+\frac{1}{2}a_tt^2$

where

d is the initial distance between the truck and the car (the truck starts some distance ahead of the car)

$a_t=2.10 m/s^2$ is the acceleration of the truck

The car position instead it is given by the equation

$x_c(t)=\frac{1}{2}a_ct^2$

where

$a_c=3.40 m/s^2$ is the acceleration of the car

The car overtakes the truck when the truck has moved 60.0 m, so when

$x_t(t') = d + 60$

Therefore, solving the equation, we find the time t when this occurs:

$d+\frac{1}{2}a_t t'^2 = d+60\\\frac{1}{2}a_tt'^2=60\\t'=\sqrt{\frac{2\cdot 60}{a_t}}=\sqrt{\frac{120}{2.1}}=7.56 s$

B)

In order to find the initial distance between the car and the truck (d), we have to calculate first the distance covered by the car during these 7.56 s. It is given by:

$x_c(t')=\frac{1}{2}a_c t'^2=\frac{1}{2}(3.40)(7.56)^2=97.2 m$

This means that after 7.56 s, when the car reaches the truck, the car has covered 97.2 m while the truck has covered 60 m. However, their positions are now equal, so we can write:

$x_c(t')=x_t(t')$

And by solving the equation, we find the value of d, the initial distance between car and truck:

$\frac{1}{2}a_c t'^2 = d + \frac{1}{2}a_t t'^2\\d=\frac{1}{2}(a_c-a_t)t'^2 = \frac{1}{2}(3.40-2.10)(7.56)^2=37.1 m$

C)

In order to find the speed of each vehicle, we use the following suvat equation:

$v=u+at$

where

u is the initial velocity

a is the acceleration

t is the time

For the truck, we have:

u = 0

$a_t = 2.10 m/s^2$

So its speed after t = 7.56 s is

$v_t = 0+(2.10)(7.56)=15.9 m/s$

For the car, we have

u = 0

$a_c=3.40 m/s^2$

So its speed after t = 7.56 s is

$v_c=0+(3.40)(7.56)=25.7 m/s$

D)

Find the graph required in attachment.

On the x-axis, it is represented the time in seconds. On the y-axis, it is represented the position in meters.

Both curves are in the shape of a parabola since the motion of both vehicles is an accelerated motion.

The curve that starts at -37.1 m is the curve representing the car: in fact, the car starts behind the truck by 37.1 m. The curve that starts from x = 0, t= 0 is that of the truck.

The two curves meets when t = 7.56 s: at that time, the two vehicles have reached the same position, and we see that occurs when x = 60 m, which means that this happens when the truck has covered 60 meters.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0

3 years ago

A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that i

mars1129 [50]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The electric field at that point is $E = 7500 \ N/C$

Explanation:

From the question we are told that

The radius of the inner circle is $r_i = 0.80 \ m$

The radius of the outer circle is $r_o = 1.20 \ m$

The charge on the spherical shell $q_n = -500nC = -500*10^{-9} \ C$

The magnitude of the point charge at the center is $q_c = + 300 nC = + 300 * 10^{-9} \ C$

The position we are considering is x = 0.60 m from the center

Generally the electric field at the distance x = 0.60 m from the center is mathematically represented as

$E = \frac{k * q_c }{x^2}$

substituting values

$E = \frac{k * q_c }{x^2}$

where k is the coulomb constant with value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

substituting values

$E = \frac{9*10^9 * 300 *10^{-9}}{0.6^2}$

$E = 7500 \ N/C$

7 0

3 years ago

Which term is a scalar quantity?

balu736 [363]

Vector quantities have both magnitude and direction. Distance is a scalar quantity. It refers only to how far an object has traveled. For example, 4 feet is a distance; it gives no information about direction.

7 0

3 years ago

Read 2 more answers

NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supp

IRINA_888 [86]

Complete question :

NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supply storage area of the lunar outpost where gravity is 1.63m/s/s can only support 1 x 10 over 5 N. What is the maximum WEIGHT of supplies, as measured on EARTH, NASA should plan on sending to the lunar outpost?

Answer:

601000 N

Explanation:

Given that :

Acceleration due to gravity at lunar outpost = 1.6m/s²

Supported Weight of supplies = 1 * 10^5 N

Acceleration due to gravity on the earth surface = 9.8m/s²

Maximum weight of supplies as measured on EARTH :

Ratio of earth gravity to lunar post gravity:

(Earth gravity / Lunar post gravity) ;

(9.8 / 1.63) = 6.01

Hence, maximum weight of supplies as measured on EARTH should be :

6.01 * (1 × 10^5)

6.01 × 10^5

= 601000 N

3 0

3 years ago

If a sled has at the top of 10 m hill had 1000 J of potential energy what would happen to the PE if the sled were to moved to a

nataly862011 [7]

Answer:

500J

Explanation:

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7 0

3 years ago