Circuit diagram showing three resistors in series with an ammeter,a cell and a switch

A car with an initial speed of 4.30 m/s accelerates at a rate of 3 ms^2. What is the car's speed after 5 s?

Lorico [155]

Answer:

v = u + at

u = 4.30 m/s

a = 3 m/s^2

t = 5 s

v = 4.30 + (3)(5)

v = 4.30 + 15

v = 19.30 m/s

Explanation:

Hope this helps!

5 0

3 years ago

ametal of mass 0.6kg is heated by an electric heater connected to 15v batter when the ammeter reading is 3A its tempeeature rise

Vedmedyk [2.9K]

Answer:

692 J/kg/°C

Explanation:

Electric energy added = amount of heat

Power × time = mass × SHC × increase in temperature

Pt = mCΔT

(15 V × 3 A) (10 min × 60 s/min) = (0.6 kg) C (85°C − 20°C)

C = 692 J/kg/°C

6 0

3 years ago

Suppose that a 710 kg car is traveling at 13 m/s. Its brakes can apply a force of 4000 N.

Maksim231197 [3]

Answer:lalalalalall

Explanation:

3 0

3 years ago

A thin rod (length = 2.97 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge.

nlexa [21]

Answer:

a) w = 2.57 rad / s , b) α = 3.3 rad / s²

Explanation:

a) Let's use the conservation of mechanical energy, we will write it in two points the highest and when touching the ground

Initial. Higher

Em₀ = U = m g h

Final. Touching the ground

$Em_{f}$ = K = ½ I w²

How energy is conserved

Em₀ = $Em_{f}$

mg h = ½ I w2

The moment of specific object inertia

I = m L²

We replace

m g h = ½ (mL²) w²

w² = 2g h / L²

The height of the object is the length of the bar

h = L

w = √ 2g / L

w = √ (2 9.8 / 2.97)

w = 2.57 rad / s

b) the angular acceleration can be found from Newton's second rotational law

τ = I α

W L = I α

mg L = (m L²) α

α = g / L

α = 9.8 / 2.97

α = 3.3 rad / s²

3 0

3 years ago

A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c

Artist 52 [7]

Answer:

Approximately $25\; {\rm N}$ (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

When the $6.0\; {\rm cm}$ -spring in this question is stretched to $10\; {\rm cm}$ , the displacement is $x = (10\; {\rm cm} - 6.0\; {\rm cm})$ .
Likewise, if this spring is stretched to $20\; {\rm cm}$ , the displacement would be $(20\; {\rm cm} - 6\; {\rm cm})$ .

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of $F_{\text{a}}$ displaces this spring by $x_{\text{a}}$ , while a force of $F_{\text{b}}$ displaces this spring by $x_{\text{b}}$ , then:

$\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}$ .

In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

$\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}$ .

7 0

2 years ago

Circuit diagram showing three resistors in series with an ammeter,a cell and a switch​

Circuit diagram showing three resistors in series with an ammeter,a cell and a switch