How far will a car travel in 4 hours at 40 mph? Use one of the following to find the answer.

A grape is thrown straight up in the air. If it was thrown at 2 m/s, how high will it go in the air?

ser-zykov [4K]

Answer:

Answer: Height is 0.204 m

Explanation:

At the highest point, it is called the maximum height.

• From third newton's equation of motion:

${ \rm{ {v}^{2} = {u}^{2} + 2gs}}$

• At maximum height, v is zero

• u is initial speed

• g is -9.8 m/s²

• s is the height

${ \rm{0 {}^{2} = {2}^{2} - (2 \times 9.8 \times s)}} \\ \\ { \rm{4 = 19.6s}} \\ \\ { \rm{s = 0.204 \: m}}$

4 0

2 years ago

Read 2 more answers

A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff

postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is 0.833.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

$F_c=\frac{mv^2}{R}$

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

$N=mg$

Therefore, frictional force, $f=\mu mg$

Now, frictional force = centripetal force

$f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}$

Plug in the given values and solve for 'μ'. This gives,

$\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833$

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0

3 years ago

A sample of gas has a volume of 215 cm3 at 23.5 °c and 84.6 kpa. what volume (cm3 will the gas occupy at stp

Dafna11 [192]

The answer is 165.3 cm³.

P1 * V1 / T1 = P2 * V2 / T2

The initial sample:
P1 = 84.6 kPa
V1 = 215 cm³
T1 = 23.5°C = 23.5 + 273 K = 296.5 K

At STP:
P2 = 101.3 kPa
V2 = ?
T2 = 273 K

Therefore:
84.6 * 215 / 296.5 = 101.3 * V2 / 273
61.34 = 101.3 * V2 / 273
V2 = 61.34 * 273 / 101.3
V2 = 165.3 cm³

6 0

3 years ago

Bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. What is the veloc

Nuetrik [128]

The velocity of B after elastic collision is 3.45m/s

This type of collision is an elastic collision and we can use a formula to solve this problem.

<h3>Elastic Collision</h3>

$v_2 = \frac{2m_1u_1}{m_1+m_2} - \frac{m_1 - m_2}{m_1 + m_2}u_2$