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emmasim [6.3K]
3 years ago
10

Add the vectors:

Physics
1 answer:
Anettt [7]3 years ago
8 0

Vector 1 has components

x_1=(10\,\mathrm m)\cos20^\circ\approx9.40\,\mathrm m

y_1=(10\,\mathrm m)\sin20^\circ\approx3.42\,\mathrm m

and vector 2 has

x_2=(10\,\mathrm m)\cos80^\circ\approx1.74\,\mathrm m

y_2=(10\,\mathrm m)\sin80^\circ\approx9.85\,\mathrm m

Add these vectors to get the resultant, which has components

x_{\rm total}\approx11.133\,\mathrm m

y_{\rm total}\approx13.268\,\mathrm m

The magnitude of the resultant is

\sqrt{{x_{\rm total}}^2+{y_{\rm total}}^2}\approx17.321\,\mathrm m

with direction \theta such that

\tan\theta=\dfrac{y_{\rm total}}{x_{\rm total}}\implies\theta\approx50^\circ

or about 50º N of E.

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How much heat is needed to melt 13.74 kg of silver that is initially at 20°C?
salantis [7]

Answer:

Q = 4.63 \times 10^6 J

Explanation:

As we know that melting point of silver is

T = 961.8 degree C

Latent heat of fusion of silver is given as

L = 111 kJ/kg

specific heat capacity of silver is given as

s = 240 J/kg C

now we will have

Q = ms\Delta T + mL

\Delta T = 961.8 - 20

\Delta T = 941.8 degree C

now from above equation

Q = (13.74)(240)(941.8) + (13.74)(111 \times 10^3)

Q = 3.1 \times 10^6 + 1.53 \times 10^6

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3 0
3 years ago
A bullet is fired horizontally from a handgun at a target 100.0 m away. If the initial speed of the bullet as it leaves the gun
Lera25 [3.4K]

Answer:

The distance is 0.53 m.

Explanation:

Given that,

Target distance = 100.0 m

Speed of bullet = 300 m/s

We need to calculate the total time

Using formula of time

t=\dfrac{d}{v}

Put the value into the formula

t=\dfrac{100.0}{300}

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Now, consider vertical motion of bullet.

Initial velocity of bullet in vertical direction = 0 m/s

We need to calculate the vertically distance

Using equation of motion

s=ut+\dfrac{1}{2}gt^2

Put the value in the equation

s=0+\dfrac{1}{2}\times9.8\times(0.33)^2

s=0.53\ m

Hence, The distance is 0.53 m.

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