All categories

4 years ago

A 1300-kg car initially has a velocity of 22.2 m/s due south. It brakes to a stop over a 180 m distance.

1 answer:

4 0

The acceleration of the object which moves from an initial step to a full halt given the distance traveled can be calculated through the equation,
d = v² / 2a
where d is distance, v is the velocity, and a is acceleration
Substituting the known values,
180 = (22.2 m/s)² / 2(a)
The value of a is equal to 1.369 m/s²
The force needed for the object to be stopped is equal to the product of the mass and the acceleration.
F = (1300 kg)(1.369 m/s²)
F = 1779.7 N

You might be interested in

A negative charge of 20 x 10-6C and another charge of 15 x 10-6C are separated by as distance of 0.7 m.

denpristay [2]

Answer:

Approximately $5.5\; \rm N$ , assuming that the volume of these two charged objects is negligible.

Explanation:

Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.

Let $q_1$ and $q_2$ denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, $q_1 = 20 \times 10^{-6}\; \rm C$ and $q_2 = 15 \times 10^{-6}\; \rm C$ .

Let $r$ denote the distance between these two point charges. In this question, $r = 0.7\; \rm m$ .

Let $k$ denote the Coulomb constant. In standard units, $k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}$ .

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\end{aligned}$ .

Substitute in the values and evaluate:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}$ .

8 0

3 years ago

A heat pump with a COP of 3.15 is used to heat an air-tight house. When running, the heat pump consumes 5 kW of power. If the te

Jet001 [13]

Answer: 1026s, 17.1m

Explanation:

Given

COP of heat pump = 3.15

Mass of air, m = 1500kg

Initial temperature, T1 = 7°C

Final temperature, T2 = 22°C

Power of the heat pump, W = 5kW

The amount of heat needed to increase temperature in the house,

Q = mcΔT

Q = 1500 * 0.718 * (22 - 7)

Q = 1077 * 15

Q = 16155

Rate at which heat is supplied to the house is

Q' = COP * W

Q' = 3.15 * 5

Q' = 15.75

Time required to raise the temperature is

Δt = Q/Q'

Δt = 16155 / 15.75

Δt = 1025.7 s

Δt ~ 1026 s

Δt ~ 17.1 min

5 0

3 years ago

Read 2 more answers

Why does the moon appear in the daytime

Brut [27]

Answer:

You can see the moon when it is in the perfect spot and it's reflecting the right amount of light. :)

Hope this helps! If you don't mind, please mark as brainliest! Thx!

4 0

3 years ago

A person looks across a stadium at the scoreboard, which appears blurry. The person looks down at the program, which is complete

Pepsi [2]

Answer is D. Nearsightedness is when a person can see near, but not far. Everything appears blurry from far away, but as you get close to it, it becomes more focused

3 0

3 years ago

Read 2 more answers

What are the two types of physical fitness?

Marina86 [1]

your question: 

What are the two types of physical fitness?

answer:

The two types of physical fitness are Health-related physical fitness and Performance-related physical fitness.

4 0

3 years ago