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3. What is the density of a substance that weighs 39.1 g and is 56.0 mL? (Density = g/mL)

Alex Ar [27]

.689214 g/ml.
Truthfully this is an easy calculation you should have been able to do but I’ll help you out

3 0

3 years ago

A projectile is launched at an angle 60° to the horizontal

Katena32 [7]

Explanation:

We have,

A projectile is launched at an angle 60° to the horizontal with some initial speed 45 m/s.

The cliff is 265 m high.

(A) The final velocity of the projectile is given by :

$v_f^2-v_i^2=2gs\\\\v_f^2=2gs+v_i^2\\\\v_f^2=2\times 9.8\times 265+45^2\\\\v_f=84.96\ m/s$

(B) The maximum height of the projectile is given by :

$H=\dfrac{v_i^2\sin^2\theta}{2g}\\\\H=\dfrac{45^2\times \sin^2(60)}{2\times 9.8}\\\\H=77.48\ m$

(C) The horizontal range of the motion is given by :

$R=\dfrac{v_i^2\sin 2\theta}{g}\\\\R=\dfrac{(45)^2\times \sin 2(60)}{9.8}\\\\R=178.94\ m$

7 0

4 years ago

A small wooden block with mass 0.750 kg is suspended from the lower end of a light cord that is 1.72 m long. The block is initia

Ierofanga [76]

Answer:

$v_{0}=319.2 m/s$

Explanation:

We need to use the momentum and energy conservation.

$p_{0}}=p_{f}$

$mv_{0}=(m+M)V_{1}$

Where:

m is the mass of bullet (m=0.01 kg)
M is the mass of the wooden (M=0.75 kg)
v(0) initial velocity of bullet
V(1) is the velocity of the combined object in the moment the bullet hist the block

Conservation of energy.

We have kinetic energy at first and kinetic and potential energy at the end.

$(1/2)(m+M)V_{1}^{2}=(1/2)(m+M)V_{2}^{2}+(m+M)gh$

Here:

V(1) is the velocity of the combined at the initial position
h is the vertical height ( h = 0.800 m)

We can find V(2) using the definition of force at this point:

$\Sigma F=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)$

$T-(m+M)gcos(\theta)=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)$

$cos(\theta) =(L-h)/L=(1.72-0.8)/1.72$

$\theta =57.66$

Now, we can solve the equation to find V(2)

$V_{2}=\sqrt{\frac{R*(T-(m+M)*g*cos(\theta))}{(m+M)}}$

$V_{2}=\sqrt{\frac{1.72*(4.86-(0.01+0.75)*9.81*cos(57.66))}{(0.01+0.75)}}$

$V_{2}=1.40 m/s$

Now we can find V(1) using the conservation of energy equation

$(1/2)V_{1}^{2}=(1/2)V_{2}^{2}+gh$

$V_{1}=\sqrt{V_{2}^{2}+2gh}$

$V_{1}=\sqrt{1.40^{2}+2*9.81*0.8}$

$V_{1}=4.20 m/s$

Finally, using the momentum equation we find v(0)

$v_{0}=\frac{(m+M)V_{1}}{m}$

$v_{0}=\frac{(0.01+0.75)*4.20}{0.01}$

$v_{0}=319.2 m/s$

I hope it helps you!

7 0

4 years ago

Where are the three seismographs used to find the epicenter of this earthquake located?

iogann1982 [59]

The seismographs are located in Minneapolis, Detroit, and Charleston.

3 0

3 years ago

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How many groups are in the modern periodic table? 12 18 22 24

Elan Coil [88]

Answer:

18

Explanation:

The s-, p-, and d-block elements of the periodic table are arranged into 18 numbered columns, or groups.

3 0

4 years ago

Read 2 more answers