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Alika [10]
2 years ago
15

A vehicle is moving with 20m/s towards the east and another is moving 15m/s towards the west​

Physics
1 answer:
just olya [345]2 years ago
6 0

Answer:

5 m/s

Explanation:

Given that,

A vehicle is moving with 20m/s towards the east and another is moving 15m/s towards the west​.

It is assumed to find the resultant velocity of the vehicle. Let east side is positive and west is negative. So,

v=v_1+v_2\\\\=20+(-15)\\\\=5\ m/s

Hence, the resultant velocity of the vehicle is equal to 5 m/s.

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Shawn and his bike have a total mass of
Trava [24]

Answer:

Shawn's kinetic energy is 61.45 J.

Explanation:

Given;

Mass of the system is, m=52.3\ kg

Displacement of the bike is, d=1.6\ km=1.6\times 1000=1600\ m

Time taken is, t=17.4\ min=17.4\times 60=1044\ s

Let the constant velocity be v.

For constant velocity, magnitude of velocity is given as distance by time.

Therefore, v=\frac{d}{t}=\frac{1600}{1044}=1.533\ m/s

Now, kinetic energy of a body is given as:

KE=\frac{1}{2}mv^2

Here, m=52.3\ kg,\ v=1.533\ m/s

\therefore KE=\frac{1}{2}\times 52.3\times (1.533)^2\\KE=0.5\times 52.3\times 2.35\\KE=61.45\ J

Therefore, Shawn's kinetic energy is 61.45 J.

3 0
3 years ago
If i punch my self and it will hurt am i weak or strong
Natasha2012 [34]

Answer:

You're strong.

Explanation:

I've been thinking of this for quite a while, and I realized that your body has a certain limit to how much pain it can take. So, punching yourself extremely hard will cause pain, because that's your body's reaction to immense pressure being put on it. But, the fact that you punched yourself so hard that it hurts, shows that you are capable of applying so much pressure; therefore, you are strong.

6 0
3 years ago
Two point charges are 10.0cm apart and have charges of 2.0uC and -2.0uC, respectively. What is the magnitude of the electric fie
elena-14-01-66 [18.8K]
The electric field generated by a point charge is given by:
E= k_e \frac{Q}{r^2}
where
k_e = 8.99 \cdot 10^9 Nm^2 C^{-2} is the Coulomb's constant
Q is the charge
r is the distance from the charge

We want to know the net electric field at the midpoint between the two charges, so at a distance of r=5.0 cm=0.05 m from each of them. 

Let's calculate first the electric field generated by the positive charge at that point:
E_1=k_e  \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =+7.19 \cdot 10^6 N/C
where the positive sign means its direction is away from the charge.

while the electric field generated by the negative charge is:
E_2=k_e \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(-2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =-7.19 \cdot 10^6 N/C
where the negative sign means its direction is toward the charge.

If we assume that the positive charge is on the left and the negative charge is on the right, we see that E1 is directed to the right, and E2 is directed to the right as well. This means that the net electric field at the midpoint between the two charges is just the sum of the two fields:
E_{tot} =E_1 + E_2 = 7.19 \cdot 10^6 N/C+7.19 \cdot 10^6 N/C=1.44 \cdot 10^7 N/C
3 0
3 years ago
Two particles, each of mass 7.0 kg, are a distance 3.0 m apart. To bring a third particle, with mass 21 kg, from far away to a r
garri49 [273]

Answer: the external agent must do work equal to -1.3 × 10⁻⁸ J

Explanation:

Given that;

Mass M1 = 7.0 kg

r = 3.0/2 m = 1.5 m

Mass M2 = 21 kg

we know that G = 6.67 × 10⁻¹¹ N.m²/kg²

work done by an external agent W = -2GM2M1 / r

so we substitute

W = (-2 × 6.67 × 10⁻¹¹ × 21 × 7) / 1.5

W = -1.96098 × 10⁻⁸ / 1.5

W = -1.3 × 10⁻⁸ J

Therefore the external agent must do work equal to -1.3 × 10⁻⁸ J

8 0
3 years ago
It is recommended that adults should get at least 150 minutes of moderate-intensity physical activity each week. they should als
shepuryov [24]
I think it's three days. I read it in assignment potion before but it's kinds fuzzy but I believe it's three days. Hopefully thats correct.
4 0
3 years ago
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