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2 years ago

Can someone help me with these questions please?

Physics

1 answer:

Alenkinab [10]2 years ago

8 0

Answer:

pang anong grade po yan kasi grade 4 palang po

kasi ako

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What is the volume of an object that has a density of 65g/cm3 and a mass of 130g.

lora16 [44]

Density ρ is mass m per unit volume v, or

ρ = m / v

Solving for v gives

v = m / ρ

So the given object has a volume of

v = (130 g) / (65 g/cm³) = 2 cm³

5 0

3 years ago

What does m/s/s mean?

Pani-rosa [81]

Explanation:

There are two answers

m/(s/s)=m

(m/s)/s=m/s²

5 0

3 years ago

An aluminum wire with a diameter of 0.115 mm has a uniform electric field of 0.235 V/m imposed along its entire length. The temp

tekilochka [14]

Answer with Explanation:

We are given that

Diameter of coil=d=0.115mm

Radius, r= $\frac{d}{2}=\frac{0.115}{2}=0.0575mm=0.0575\times 10^{-3} m$

Using $1mm=10^{-3} m$

Electric field=E=0.235V/m

T=55 degree C

$T_0=20^{\circ} C$

$\rho_0=2.82\times 10^{-8}\Omega m$

$\alpha=3.9\times 10^{-3}/C$

(a).We know that

$\rho=\rho_0(1+\alpha(T-T_0))$

Substitute the values

$\rho=2.82\times 10^{-8}(1+3.9\times 10^{-3}(55-20))$

$\rho=3.2\times 10^{-8}\Omega m$

(b).Current density, $J=\frac{E}{\rho}$

Using the formula

$J=\frac{0.235}{3.2\times 10^{-8}}=7.3\times 10^6A/m^2$

c.Total current,I=JA

Where $A=\pi r^2$

$\pi=3.14$

Using the formula

$I=7.3\times 10^6\times 3.14\times (0.0575\times 10^{-3})^2$

I=0.076A

d.Length of wire=l=2m

$V=El$

Substitute the values

$V=0.235\times 2=0.47 V$

5 0

3 years ago

Please help formula to calculate equivalent resistance

azamat

Answer:

(Equation 10.3. 2): RS=R1+R2+R3+R4+R5=20Ω+20Ω+20Ω+20Ω+10Ω=90Ω

6 0

3 years ago

a 282 kg bumper car moving right at 3.50 m/s collides with a 155 kg bumper car moving 1.88 m/s left. afterwards, the 282 kg car

Vaselesa [24]

The momentum of the 155 kg car afterwards is 469.7 kg m/s to the right

Explanation:

We can solve the problem by using the law of conservation of momentum: the total momentum of the system is conserved before and after the collision, so we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 282 kg$ is the mass of the bumper car

$u_1 = 3.50 m/s$ is the initial velocity of the bumper car (we take the right as positive direction)

$v_1 = 0.800 m/s$ is the final velocity of the bumper car

$m_2 = 155 kg$ is the mass of the second bumper car

$u_2 = -1.88 m/s$ is the initial velocity of the second car (moving to the left)

$v_2$ is the final velocity of the second car

Solving for $v_2$ ,

$v_2 = \frac{m_1 u_1+m_2 u_2 - m_1 v_1}{m_2}=\frac{(282)(3.50)+(155)(-1.88)-(282)(0.800)}{155}=3.03 m/s$

where the positive sign means the direction is to the right.

And now we can find the momentum of the 155 kg afterwards, which is

$p_2 = m_2 v_2 = (155)(3.03)=469.7 kg m/s$ (to the right)

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

7 0

3 years ago